/ 


-^  —  tbfi  last  date  stamped  be' 


SITY  OF  CAUFGi 

LIBRARY, 

1«S  ANGELES,  GAL  IF* 


APPLICATIONS  OF  THE 
CALCULUS  TO  MECHANICS 


BY  ^a^ 

K  E.  HEDRICK 


Pbofessob  of  Mathematics  in  the  University  of  Missodri 


t^ 


).  D.  KELLOGG 

Assistant  Professor  of  Mathematics  in  the  University  of  Missouri 


GINN  AND  COMPANY 

BOSTON  •  NEW  YORK  •   CHICAGO  ■  LONDON 


Copyright,  1909,  by 
E.  R.  Hedrick  and  O.  D.  Kellogg 


ALL  BIGHTS  RESERVED 
89.11 


Cfae   gtftenttutn   jgretfg 

GINN  AND  COMPANY  ■  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


; 


f,( 


QA        Sciences 
H2>5 


PEEFACE 

Wlierever  the  teaching  of  mathematics  to  engineering  students 
is  discussed,  and  frequently  in  cases  of  other  classes  of  students, 
the  criticism  which  is  almost  without  exception  the  most  insistent 
is  this :  that  the  student  leaves  the  course  without  adequate  abihty 
to  apply  his  mathematical  knowledge.*  This  means  that  he  has 
not  the  faculty  of  taking  a  problem,  giving  it  an  analytic  formu- 
lation, and  interpreting  the  analytic  results.  It  is  an  open  question 
whether  it  is  the  duty  of  the  teacher  of  mathematics,  or  of  the 
teacher  of  the  more  technical  work  which  involves  mathematics, 
to  supply  the  needed  training,  but  usually  the  mathematician  is 
glad  at  least  to  share  the  responsibility  and  to  do  whatever  he  can 
to  make  his  work  fruitful,  fully  conscious  of  the  fact  that  if  he 
can  successfully  make  the  contact  of  his  subject  with  the  problems 
of  the  laboratory,  of  the  engineering  office,  and  of  other  activities, 
he  will  thereby  add  immensely  to  the  vitality  and  interest  of  his 
work.  With  such  a  motive,  it  has  been  the  practice  at  the  Uni- 
versity of  Missouri  to  follow  the  course  in  sophomore  calculus 
with  several  weeks  in  applications  to  mechanics,  this  being  a  sub- 
ject rich  in  the  kind  of  material  desired.  The  present  book  is  a 
formulation  of  the  work  there  attempted,  and  it  is  believed  that 
the  need  at  our  institution  which  has  called  the  book  into  being 
will  make  its  appearance  welcome  to  a  large  number  of  mathe- 
matical departments. 

Opinions  will  differ  as  to  the  subject-matter  which  such  a  book 
should  contain.  The  authors  were  guided  by  the  feeling  that  it 
was  practice  in  applying  calculus  rather  than  a  broad  knowledge 

*  See,  for  instance,  the  reports  of  the  joint  meeting  of  mathematicians  and 
engineers  held  in  Chicago,  December,  1907,  under  the  auspices  of  the  Chicago  Section 
of  the  American  Mathematical  Society.  These  reports  appeared  in  Science  during 
the  ensuing  year. 

iii 


iv  PREFACE 

of  mechanics  that  was  desired,  and  that  such  an  end  would  be 
hindered  rather  than  helped  by  a  wide  diversity  of  subject-matter. 
It  was  felt  that,  when  feasible,  the  student  secures  a  better  insight 
into  a  subject  by  developing  a  portion  of  the  theory  himself,  and 
so  "  exercises  "  have  been  introduced  which  form  a  part  of  such 
development  of  the  theory.  The  "  problems  "  are  the  applications 
of  the  theory,  usually  to  cases  in  which  numerical  data  are  given. 
They  vary  considerably  in  difficulty,  but  it  is  thought  that  they 
are  sufficiently  numerous  to  supply  a  good  number  of  the  proper 
grade  for  any  given  class ;  and,  furthermore,  it  is  believed  that 
a  more  difficult  problem  with  a  proper  amount  of  elucidation  in 
advance  by  the  teacher  will  be  of  far  more  value  to  the  student 
than  a  number  of  problems  so  well  within  his  range  of  ability  as 
to  require  very  little  study  on  his  part  concerning  the  method 
of  attack.  Some  hints  will  be  found  in  the  text,  and  some  sug- 
gestions are  given  among  the  answers  at  the  end  of  the  book. 

We  have  endeavored,  by  a  judicious  selection  of  the  problems 
to  which  answers  are  given,  to  secure  a  safe  mean  between  the 
evil  of  supplying  detailed  answers,  which  rob  the  student  of  his 
independence,  and  the  evil  of  furnishing  him  with  no  check  upon 
liis  work.  We  shall  be  most  grateful  for  any  corrections  or  sug- 
gestions concerning  tliis  or  other  aspects  of  the  book. 

E.  R.  HEDRICK 
Columbia,  Missodri  O.  D.  KELLOGG 


CONTENTS 


CHAPTER  I.    INTRODUCTION 

SECTION  PAGE 

1.  Mechanics.    Subject-matter  of  mechanics 1 

2.  Units 1 

3.  Vectors 3 

4.  The  fixing  of  vectors 3 

5.  Operations  upon  vectors 4 

6.  The   fundamental   relation   between   acceleration   and   force 

vectors 8 

Analysis  of  Chapter  I 11 

CHAPTER  II.    STATICS 

7.  Statics.    Subject-matter  of  statics 12 

8.  Equilibrium  under  the  action  of  concurrent  forces     ....  12 

9.  Nonconcurrent  forces  ;  moments 16 

10.  Couples 18 

11.  Equilibrium.     General  conditions  for  equilibrium  of  a  rigid 

body 20 

12.  Centers  of  mass  of  systems  of  particles 26 

13.  Centers  of  mass  of  continuous  bodies 28 

11.   Centers  of  mass  of  some  bodies  of  special  shape;       ....  30 

Analysis  of  Chapter  II    ....  ' 41 

CHAPTER  III.    DYNAMICS  OF  A  PARTICLE 

15.  Dynamics 42 

16.  Rectilinear  motion  ;  concepts  involved 42 

17.  Some  special  rectilinear  motions 48 

18.  Motion  of  a  particle  in  a  plane 55 

19.  Some  geometric  properties  of  plane  motion 57 

20.  The  differential  equations  of  plane  motion 63 

21.  Some  special  plane  motions 63 

Analysis  of  Chapter  III 77 

v 


vi  CONTENTS 

CHAPTER  IV.    WORK  AND  ENERGY 

SECTION  PAGE 

22.  Work 78 

23.  Force  fields  ;  work  on  curved  paths 79 

24.  Conservative  fields  ;  potential  energy 80 

25.  Kinetic  energy 82 

26.  Conservation  of  energy 83 

Analysis  of  Chapter  IV 85 

CHAPTER  V.    MECHANICS  OF  RIGID  BODIES 

27.  Instantaneous  motion  of  a  rigid  body 86 

28.  Pure  translation 89 

29.  Pure  rotation 90 

30.  Moments  and  products  of  inertia.    Radius  of  gyration  ...  92 

31.  Work  and  energy  in  the  case  of  a  rigid  body 98 

32.  Compound  pendulum  ;    experimental  determination  of   mo- 

ments of  inertia 102 

33.  General  equations  of  motion  of  a  rigid  body 104 

Analysis  of  Chapter  V 107 

SUGGESTIONS  AND  ANSWERS 109 

INDEX 115 


APPLICATION'S  OP 
THE  CALCULUS  TO  MECHAOTCS 


CHAPTER  I 

INTRODUCTION 

1.  Mechanics.  Mechanics  deals  with  the  position  or  motion  of 
bodies  in  space ;  so  that  the  ideas  of  space  and  time  implied  in 
motion,  and  of  mass  implied  in  body,  are  fundamental.  These 
three  concepts  cannot  be  defined,  for  there  is  nothing  simpler  in 
terms  of  which  we  can  define  them. 

2.  Units.  With  each  of  the  three  concepts  mentioned  is  asso- 
ciated the  idea  of  quantity,  the  quantity  being  measured  by  com- 
parison with  a  conventional  unit  or  standard  quantity.  In  Great 
Britain,  and  for  commercial  purposes  in  the  United  States,  the 
units  of  space,  mass,  and  time  generally  employed  are  the  foot, 
pound,  and  second  respectively.  The  initials  of  these  fundamental 
units  are  usually  used  to  designate  the  system,  "  the  F.P.S.  system." 
In  France,  and  for  scientific  purposes  in  the  United  States,  the 
units  are  the  centimeter,  gram,  and  second  respectively,  giving 
"the  C.G.S.  system."  The  following  table  gives  the  important 
derived  units  of  velocity,  acceleration,  and  force.  In  engineering 
circles  other  units  of  mass  are  in  use,  with  the  result  that  the 
units  of  force  are  also  changed.  The  "  engineering "  or  "  tech- 
nical" or  "gravitational"  units  of  mass  and  force  are  also  given 
below.* 

*  For  further  information  concerning  units,  see  Ziwet,  Tlieoretical  Mechanics; 
JSncyclopedia  Britannica  on  "Weights  and  Measures."  In  the  following  we  shall 
meet  with  other  derived  units,  hut  these  will  he  defined  as  they  arise. 

1 


INTRODUCTION 


Quantity 


Velocity  =  time- 
rate  of  change 
of  space 


British  ok  F.P.S.  System 


One  foot  per  second. 


French  or  C.G.S.  System 


One  centimeter  per  second. 


Acceleration  = 
time-rate  of 
change  of  ve- 
locity   .     .     . 


One  foot  per  second  per 
second.  (The  accelera- 
tion of  a  falling  body 
near  the  earth's  surface 
is  ^  =  32.2  of  these  units, 
nearly.) 


One  centimeter  })er  second 
per  second.  (The  accelera- 
tion   of    a    falling    body 
near   the  earth's  surface  / 
is  g  z=  981  of  these  units,/ 
nearly.)  / 


Force  =  Mass  x 
Acceleration 


One  poundal :  the  force 
which,  acting  constantly 
throughout  a  second,  will 
give  to  a  pound  of  matter 
a  unit  acceleration,  that 
is,  increase  the  velocity 
by  one  foot  jier  second. 


One  dyne :  the  force  which, 
acting  constantly  through- 
out a  second,  will  give  to 
a  gram  of  matter  a  unit 
acceleration,  that  is,  in- 
crease the  velocity  by  one 
centimeter  per  second. 


The  engineeriug  units  of  mass  and  force  in  the  twoy  systems 
follow : 


QlTAXTITY 

British  ok  F.P.S.  System 

French  or  C.Q.S.  System 

Mass     .... 

The  mass  of  a  body  weigh- 
ing g  pounds.    (To  deter- 
mine the  mass  of  a  body 
in  technical  units,  divide 
its  weight  in  pounds  by 
flr  =  32.2.) 

The  mass  of  a  body  weigh- 
ing g  kilograms.  (To  de- 
termine the  mass  of  a  body 
in  technical  units,  divide 
its  weight  in  kilograms 
by.<7=981.) 

Force  =  Mass  x 
Acceleration 

One  pound  :    the  force 
which  the  earth  exerts  at 
its  surface  upon  a  body 
weighing    one    pound. 
(One   pound  =  g  pound- 
als.) 

One  kilogram  :  the  force 
which  the  earth  exerts  at 
its  surface  upon  a  body 
weighing  one  kilogram. 
(One  kUogram  =  1000  g 
dynes.) 

One  foot      =  30. .5  centimeters. 
One  pound  =  .373  kilograms. 


One  centimeter  =  .0328  feet. 
One  kilogram     =  2.68  pounds. 


VECTOES  3 

I.   PROBLEMS   ON  UNITS 

1.  A  force  equal  to  the  "weight  of  30  gm.  acts  upon  a  mass  of  2  kgm. 
What  is  the  acceleration  in  C.G.S.  and  in  F.P.S.  units? 

Solution.  The  equation  /=  ma  holds  if  f  is  measured  in  dynes  and  vi  in 
grams.  It  may  be  solved  for  a,  thus  giving  the  desired  result.  The  weight 
of  30  gm.  is  a  force  of  30  g  dynes,  and  2  kgm.  equals  2000  gm.  Hence 
30  (7  =  2000  rt,  and  a  =  30  5r/2000  =  30  x  981/2000,  which  is  approxi- 
mately 14.7  cm.  per  sec.  per  sec.  As  one  centimeter  is  .0328  in.,  this  is 
14.7  X  .0328,  or  .482  ft.  per  sec.  per  sec. 

2.  A  man  walks  5  mi.  per  hour.  What  is  his  speed  in  F.P.S.  units, 
and  what  in  C.G.S.  units? 

3.  A  mass  of  25  gm.  moves  with  an  acceleration  of  30  cm.  per  sec. 
per  sec.  What  is  the  force  acting,  measured  in  C.G.S.  and  in  F.P.S. 
units  ? 

4.  If  sound  travels  1000  ft.  per  sec.  in  air,  what  is  its  speed  measured 
in  C.G.S.  units? 

5.  A  boy  throws  a  ball  with  a  velocity  of  50  m.  per  sec.  W^hat  is  its 
speed  in  F.P.S.  units? 

6.  A  force  of  6  lb.  acts  upon  a  3-lb.  weight.  What  is  the  acceleration 
in  F.P.S.  and  C.G.S.  units? 

3.  Vectors.  The  units  above  apply  primarily  to  bodies  moving 
in  a  straight  line.  In  the  case  of  bodies  not  moving  in  a  straight 
line,  velocity,  acceleration,  and  force  cannot  be  characterized  by 
one  measurement  alone,  say  of  their  magnitude ;  but  in  addition 
their  direction  must  be  specified.  They  are  examples  of  what  are 
called  vectors,  and  are  usually  represented  geometrically  by  directed 
straight  line  segments. 

4.  The  fixing  of  vectors.  For  the  fixing  of  vectors  in  space 
three  magnitudes  are  necessary.  These  may  be  selected  in  various 
ways ;  we  mention  here  only  two  of  them.  Let  us  denote  the 
vector  by  V.  It  is  to  be  noted  then  that  V  does  not  stand  for  a 
number  in  the  ordinary  sense,  but  for  a  set  of  numbers.  The  first 
way  of  fixing  V  is  by  its  magnitude,  v,  essentially  a  positive  num- 
ber, and  the  angles  a,  /3,  and  7  which  it  makes  with  the  coordinate 
axes  (see  Fig.  1).  These  three  angles  are  not  independent,  for 
cos^a  -f  cos^/S  -J-  cos^7  =  1 ;  so  that  V  involves  but  three  independ- 
ent magnitudes,  v  and  two  of  the  three  angles  a,  /S,  and  7. 


4  INTRODUCTION 

The  relation  given  follows  from  the  law  for  the  diagonal  of  a  cuboid 
tr*  =  y2  ^  j^.2  _^  f.2^  whence,  dividing  by  v'^,  we  have 

1  =  (f'x/'-O^  +  ('V/'O^  +  (l'z/0^  =  ^^OS^tr  +  COS2/3  +  cos2y. 


Fig.  1 

Another  means  of  fixing  V,  and  that  usually  used  in  analytic 
treatments  of  vectors,  is  by  its  projections  on  the  axes,  v^,  v^,  v^* 
If  the  initial  point  of  the  vector  be  placed  at  the  origin,  these  pro- 
jections are  simply  the  coordinates  of  the  terminal  points. 

Ex.  1.  Express  r^.,  v^,  v^  in  terms  of  v,  a,  f3,  and  y,  and  conversely. 

Ex.  2.  If  our  vector  is  required  to  lie  in  a  fixed  plane,  two  quantities 
suffice  to  fix  it.  Give  two  pairs  of  quantities  analogous  to  the  two  sets 
given  above  for  space,  and  give  the  relations  between  them. 

5.  Operations  upon  vectors.  If  an  object  move  from  a  point  A 
on  a  table  to  a  point  B  on  the  table,  while  at  the  same  time  the 
table  moves  so  that  the  point  B  passes  to  a  point  C  fixed  in  the 
room,  the  body  originally  at  A  will  arrive  at  C,  and  this  no  matter 

*  If  I  stands  for  a  given  direction,  Vj  stands  for  the  projection  of  the  vector  V 
upon  a  line  with  that  direction. 


VECTOR  OPERATIONS  5 

how  the  separate  motions  are  carried  out.  Now  the  motion  of  the 
body  relative  to  the  table  may  be  represented  by  a  vector  F^,  while 
the  motion  of  the  table  may  be  represented  by  a  vector  V^,  and  the 
total  effect  also  amounts  to  a  vector  V^.  The  last  is  said  to  be  the 
Slim  of  the  first  two,  and  this  illustrates  the  general  definition : 
place  the  initial  point  of  one  vector  upon  the  terminal  point  of  the 
other  ;  the  vector  running  from  the  free  initial  point  to  the  free 
terminal  point  is  called  the  sum,  or  resultant  vector. 

Ex.  1.  Show  that  Fg  will  be  the  same  whether  the  initial  point  of  V^ 
be  put  upon  the  tei-niinal  point  of  Fj  or  the  initial  point  of  F^  be  put  upon 
the  terminal  point  of  V^. 

Ex.  2.  Show  that  another  method  of  adding  vectors,  equivalent  to  the 
above,  is  to  put  the  initial  points  of  Fj  and  V^  together,  to  complete  the 
parallelogram,  and  to  draw  the  diagonal  from  the  common  initial  point. 
This  diagonal  will  be  V^. 

Ex.  3.  How  would  you  obtain  the  sum  of  3  vectors?  of  any  given 
number  of  vectors?    Show  that  n  vectors  may  be  added  in  any  order. 

On  the  other  hand,  a  given  vector  may  be  resolved  into  two 
vectors  whose  sum  or  resultant  it  is.  This  may  be  done  in  an 
unlimited  number  of  ways.  Several  especially  important  general 
ways  are  indicated  in  the  following  exercises. 

Ex.  4.  Let  there  be  given  in  a  plane  a  vector  F  and  two  nonparallel 
lines  /j  and  l^.  Show  that  there  are  two  vectors,  F^  parallel  to  Z^  and  V^ 
parallel  to  l^,  whose  sum  is  F.  Give  the  construction  for  V^  and  V^-  Is 
there  more  than  one  solution  ? 

Ex.  5.  Let  there  be  given  in  space  a  vector  F  and  three  lines  Zj^,  l.^,  and 
Zg,  not  in  the  same  plane  or  parallel.  Show  that  there  are  three  vectors, 
one  parallel  to  Zj,  one  to  l^,  and  one  to  Zg,  whose  sum  is  F.  Is  there  more 
than  one  solution?  To  get  a  good  figure,  draw  parallels  to  Z^,  Zg,  and  Zg 
through  the  origin  of  F,  so  that  one  looks  into  the  solid  angle  formed. 

Ex.  6.  Let  there  be 
given  a  vector  F  and  a  ^^^^^^^  ^~~''--~.  Vi 

vector  Fj.  Show  how  to 
construct  a  vector  F^ 
such  that  Vi  +  Vr^  =  V.'' 

Ex.  7.  Let    Fj    and 
Fo  be  two  vectors  whose 


^'i 


resultant  is  F,  and  let  ^10   2 

their  magnitudes  be  v^, 

Vg,  and  V  respectively;  let  a^  and  a^  be  the  positive  angles  between Fj^  and 

F  and  between  V^  and  F  respectively ; 


6  INTKODUCTION 

(a)  Show  that  the  magnitude  v  of  the  resultant  is  the  square  root  of  the 
sum  of  the  squares  of  the  magnitudes  of  the  components  increased  by  twice 
their  product  times  the  cosine  of  the  angle  (a^  +  a^  between  them,  that  is, 
t;2  =  v^  +  i?|  +  2  v^v^  cos  (a^  +  a^). 

(b)  Show  that  sin  Oj/cj  =  sin  rt2/i'i. 

(c)  Show  that  tana^  =  [v^  sin(a^  +  a^)2/[v^  +  v^  cos(aj  +  o.^)"]- 

Hint.  The  first  two  parts  are  essentially  formulas  from  plane  trigonom- 
etry. The  third  should  be  obtained  from  the  figure  by  dropping  a  perpen- 
dicular from  the  tip  of  V  to  F^  produced.  These  results  are  useful  in  finding 
direction  and  magnitude  of  resultants. 

Ex.  8.  Take  coordinate  axes  so  that  V  lies  on  the  x-axis  beginning  with 
the  origin.  Find  the  projections  on  the  axes  of  all  the  vectors,  and  thus 
prove  analytically  by  the  methods  of  Analytic  Geometry  the  formulas  of 
Ex.  7,  (a)  and  (b). 

Hint.  Show  that  v  =^  v^cos  a^  + v^cos  a^,  0  =  i'^  sin  a^  —  f 2  sin  Cg  ;  then 
square  and  add. 

It  should  be  noted  that  in  the  above  statements  concerning  com- 
position and  resolution  of  vectors  it  must  be  possible  to  regard 
the  vectors  as  acting  at  the  same  point,  or  as  concurrent.  The  fol- 
lowing important  facts  concerning  operations  with  vectors  should 
be  clearly  understood  by  the  student. 

I.  TJie  projections  upon  each  coordinate  axis  of  the  sum  of  two 
vectors  is  the  sum  of  the  projections  upon  that  axis  of  the  two  cotn- 
ponent  vectors.    The  same  is  true  for  the  projections  upon  any  line. 

Ex.  9.  Prove  the  above  theorem,  making  use  of  the  "  projection  "  the- 
orem of  the  Trigonometry  or  Analytic  Geometry  text-books.  Draw  your 
own  figure. 

Ex.  10.  Extend  the  above  theorem  to  n  vectors  (^j,  F^,  Z.^,  (Xj,  Y^, 
Z^),  •••,  (X„,  Y„,  Z„),  with  resultant  (X,  Y,  Z).  That  is,  show  that 
X  =  A\  +  X^  +  Xg  +  ---  +X„,  etc. 

Such  a  sum  is  usually  abbreviated  in  mathematical  writing  by  2A'j,  the 
Greek  sigma,  2,  denoting  a  sum. 

II.  The  projections  on  each  axis  of  k  times  a  vector,  where  k  is  a 

number,  is  k  ti7nes  the  projection  upon  the  axis  in  question  of  the 

vector.    The  same  is  true  of  the  projection  upon  any  line. 

Ex.  11.  This  amounts  to  a  definition  of  product  of  a  vector  hy  a  number. 
Verify  the  fact  that  it  holds  when  k  is  a  positive  integer  and  the  multipli- 
cation is  regarded  as  repeated  addition.  Show  that  in  all  cases  the  product 
vector  has  the  same  direction  as  the  given  one,  and  that  its  magnitude  is 
k  times  that  of  the  given  one. 


VECTOR  OPERATIONS  7 

If  a  vector  vary  with  the  time  t,  or  any  other  parameter,  its 
derivative  may  be  defined,  for  we  know  how  to  subtract  vectors 
and  to  divide  by  numbers.  The  notion  of  limit  of  a  set  of  vectors 
will  be  sufficiently  clear.  The  derivative  will  be  found  to  have 
the  property : 

III.  The  projection  upon  each  axis  of  the  derivative  of  a  vector 
is  the  derivative  of  the  projection  upon  the  axes  in  question  of  the 
vector.    The  same  is  trite  of  the  projection  upon  any  line. 


i 

dv       ^  ^ 

Fig.  3 

Ex.  12.  Working  in  two  dimensions,  draw  the  variable  vector  V  for 
various  A'alues  of  <,  approaching  <q,  and  construct  the  ajipropriate  difference 
vectors,  multiplying  their  length  by  \/td.  The  set  of  vectors  thus  obtained 
should  have  a  limiting  position,  which  we  call  the  derivative  vector.  Find 
the  magnitudes  of  the  various  approximating  vectors  and  show  that  the 
limit  of  these  magnitudes  is  •sJ(dv^/dtY  +  (do^/dt)^.  Next  show  that  the 
limit  of  their  slojie  is  (dn^^/dt)  h-  (dv^/dt).  Then  noting  that  these  are 
the  magnitude  and  slopes  of  the  vector  whose  projections  are  the  derivatives 
of  the  projections  of  V,  the  truth  of  the  above  statement,  III,  is  apparent. 

It  is  very  important  to  note  that  the  derivative  vector  in  general 
will  have  a  different  direction  from  the  given  vector  (see  p.  57). 

We  next  consider  a  very  useful  question,  namely,,  given  the  pro- 
jections of  a  vector  on  the  coordinate  axes,  supposed  rectangular, 
to  find  the  projections  upon  any  given  line.  We  consider  the 
problem  for  vectors  in  a  plane,  where  we  shall  need  it  most,  and 


8 


INTRODUCTION 


leave  the  analogous  considerations  in  space  to  the  student  to  work 
out,  or  look  up,  as  occasion  arises.  Let  W  denote  a  vector  and  I  a 
direction,  e.g.  the  direction  of  the  positive  y-axis.  We  denote  by 
{I,  W)  the  angle  between  the  direction  I  and  W  measured  from  I 


-.(2/,0 


,'''^' 


Fig.  4 

toward  W,  and  similarly  by  (x,  I)  and  (i/,  I)  the  angle  the  line  I 
makes  with  the  axes  measured  from  the  axes  to  the  line.  Then, 
according  to  the  definition  of  projection,  the  projection  of  W 
upon  I  is 

Wf  =  w  cos  (/,  W)  =  w  cos  [{x,  W)  —  (x,  I)] 

=  w  [cos  {x,  W)  cos  {x,  I)  +  sin  {x,  W)  sin  {x,  I)] 
=  w  cos  {x,  W)  •  cos  {x,  l)+  w  cos  {y,  W)-  cos  {y,  I). 
But  w  cos  (a?,  W)  =  w^,  and  w  cos(y,  W)  =  w^.    Hence  we  have 

IV.  Wi  =  w-c  cos  {x,  I)  +  Wy  cos  (y,  /),  that  is,  the  ^projection  of  a 
vector  upon  any  line  is  the  sum  of  the  products  of  each  projection 
upon  an  axis  hy  the  corresponding  direction  cosine  of  the  line. 

6.  The  fundamental  relation  between  acceleration  and  force 
vectors.  With  the  above  information  concerning  vectors,  we  may 
proceed  to  consider  motion  other  than  in  a  straight  line  (cf.  §  3), 
Velocities,  accelerations,  and  forces  will  then  be  vectors,  and  the 
fundamental  equation  F  =  mA  is  to  be  understood :  the  vectors  F 
and  A  have  the  same  direction,  and  the  magnitude  of  F  is  m  times 
the  magnitude  of  A. 


PROBLEMS  ON  VECTORS  9 

Ex.  Show  that  the  above  statement  is  equivalent  to :  the  projection 
of  F  upon  any  line  is  m  times  the  projection  upon  the  same  line  of  A. 

n.  PROBLEMS  ON  VECTORS 

The  following  problems,  as  ^yell  as  most  of  those  in  the  book,  should  be 
worked  with  the  aid  of  a  careful  figure.  Measurement  of  the  figure  will 
then  serve. as  an  approximate  check  on  the  work.  In  case  of  vectors  in 
space  the  student  can  at  least  lay  off  the  projections  on  straight  lines. 

1.  A  sailboat  is  sailing  "  into  the  wind,"  its  course  making  an  angle  of 
35°  with  the  wind.  The  speed  of  the  boat  is  6  mi.  per  hr.  and  that  of  the 
wind  is  20  mi.  per  hr.  Find  the  speed  and  direction  of  the  wind  as  it 
appears  to  a  person  on  the  boat. 

Solution.  The  forward  velocity  of  the  boat  gives  a  relative  backward 
component  to  the  velocity  of  the  wind.  Our  problem  is  to  solve  the  tri- 
angle BA  D,  given  AD  =  20,  BA=Q, 

and  2J.BAD  =  U5°.  ^- -_^ 

JBi)2=62  +  202  +  2x6x20xcos35°  y^  ^^'^ 

=  436  +  240  X  .819  /^^^''' ' 

=  25.2,  about.  j^^^^/ 

sin  DBA  =  (20  sin  145°)/ 25.2  y^^'^  ,/' 

=  20  X  .574/25.2  /^     ,<"" 

=  .4.56.  j^^       y     \^' 

2)5.4  =27.1°,  about.  ^         ^ 

Hence  the  apparent   speed   of    the 

wind  is  25.2  mi.  an  horn-,  and  it  apparently  makes  an  angle  27.1°  with  the 
course  of  the  boat. 

In  some  problems  the  formulas  of  the  exercises  in  this  chapter  may 
prove  useful,  but  the  student  had  better  rely  upon  his  knowledge  of  trigo- 
nometry and  his  own  ingenuity. 

2.  Two  forces  of  12  and  16  lb.  respectively  act  at  an  angle  of  90°. 
Find  magnitude  and  direction  of  the  resultant. 

3.  Find  the  resultant  of  two  forces  of  the  same  magnitude  y  acting  at 
an  angle  of  45°. 

4.  Two  men  kick  a  football  at  the  same  instant.  One  kicks  eastward 
at  the  rate  of  71  ft.  per  sec,  and  the  other  northwest  at  the  rate  of  48  ft. 
per  sec.    Find  the  initial  direction  and  the  velocity  of  the  ball. 

5.  The  resultant  of  two  forces  is  9.  One  of  them  is  3  and  the  angle 
which  it  makes  with  the  resultant  is  60°.  What  is  the  magnitude  of  the 
other  force,  and  what  the  angle  between  it  and  the  resultant  ? 

6.  Two  forces  are  inclined  to  their  resultant  at  angles  of  120°.  How  are 
the  magnitudes  of  the  forces  related  ? 

7.  A  balloon  rises  1120  ft.  per  min.  while  the  wind  blows  it  horizontally 
370  ft.  per  min.  "What  is  its  velocity,  and  in  which  direction  does  it  rise? 
(Use  table  of  natural  functions.)  . 


10 


INTRODUCTION 


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VI 

PROBLEMS  ON  VECTORS  11 

10.  A  man  jumps  from  a  train  rnnning  at  10  mi.  per  hr.  in  a  direction 
making  an  angle  of  30°  with  the  train  and  a  velocity  of  9  ft.  per  sec. 
What  is  his  velocity  in  magnitude  and  direction  relative  to  the  ground  ? 

11.  A  stream  has  a  current  speed  a,  and  a  man  can  row  his  boat  with  a 
speed  b.  In  what  direction  is  he  to  row  if  he  is  to  land  at  a  point  directly 
opposite  his  starting  point?  In  what  direction  must  he  row  in  order  to 
cross  in  the  shortest  time  ?    (Consider  a  $  6.) 

12.  A  train  is  traveling  at  the  rate  of  20  mi.  per  hr.,  and  rain  falls  ver- 
tically with  a  velocity  of  22  ft.  per  sec.  Find  the  direction  of  the  splashes 
on  the  windows. 

13.  A  river  |  mi.  wide  has  a  current  of  3  mi.  per  hr.;  a  ferryboat  can 
cross  in  4.5  min.  At  what  angle  must  it  be  headed  upstream  in  order  to 
go  straight  across  ? 

ANALYSIS  OF  CHAPTER  I  * 

1.  Subject-matter  of  mechanics. 

2.  British  and  French  units  of  space,  mass,  time,  velocity,  acceleration, 
and  force. 

3.  Vectors  in  space  and  in  the  plane,  and  how  they  are  fixed. 

4.  Vector  addition  and  resolution.  Some  theorems  connecting  opera- 
tions upon  vectors  with  operations  upon  their  projections. 

5.  The  fundamental  relation  between  force  and  acceleration. 

*  The  analyses  of  the  chapters  are  intended  to  point  out  the  more  important 
notions  derived  from  the  text,  so  that  the  student  may  have  some  idea  of  the  results 
he  should  have  from  his  study.  He  should  be  able  to  give  clear  and  accurate  discus- 
sions of  each  topic  noted.  He  should,  moreover,  liave  a  clear  notion  of  how  to  attack 
each  of  the  problems,  whether  or  not  he  actually  works  them  all.  A  good  percentage 
of  them  should  be  completed. 


CHAPTER  II 

STATICS 

7.  Statics  considers  primarily  the  question,  IVJien  can  bodies 
remain  tinmoved  under  the  action  of  forces  ?  although  it  will  be 
found  that  the  necessary  conditions  for  a  state  of  rest  also  allow  of 
certain  motions,  as  uniform  motion  in  a  straight  line,  and  for  this 
reason  such  motions  are  sometimes  considered  in  statics.  We  shall 
here,  however,  think  of  our  problem  as  that  of  finding  the  conditions 
for  rest.  When  these  are  fulfilled  the  forces  are  said  to  he  in.  equi- 
librium. An  essential  distinction  arises  on  the  basis  as  to  whether 
the  forces  all  act  at  one  point  or  not,  and  we  take  up  first  the 
simpler  case  in  which  they  do,  that  is,  in  which  they  are  concurrent. 

8.  Equilibrium  under  the  action  of  concurrent  forces.  We 
shall  be  concerned  only  with  riffid  bodies,  that  is,  bodies  in  wliich 
the  forces  acting  produce  no  measurable  change  in  size  or  shape.* 
The  forces  acting  on  such  a  body  may  evidently  be  displaced  along 
their  lines  of  action  by  any  amount,  for  it  is  clear  that  the  inter- 
vening little  bars  of  matter  may  be  regarded  as  transmitters  of  the 
force.  For  our  present  purposes,  therefore,  "  concurrent  forces " 
means  forces  whose  lines  of  action  pass  through  a  common  point, 
and  we  may  confine  our  attention  to  that  point. 

From  the  fundamental  law  of  §  6  we  see  that,  inasmuch  as  for 
rest  A  =  0,  it  follows  that  F=  0,  it  being  of  course  understood 
that  F  comprises  all  the  forces  acting  on  the  body,  or  that  it  is 
their  resultant.    We  then  have  the  result 

A  body  can  be  at  rest  under  the  action  of  concurrent  forces  when, 
and  only  ivhen,  their  resultant  vanishes. 

*  Of  course  no  absolutely  rigid  bodies  exist  in  nature.  Metals  whit-h  are  iisually 
thought  of  as  rigid  become  quite  pliable  when  made  in  long  thin  pieces,  like  wires. 
However,  the  notion  of  rigid  bodies  is  extremely  useful  iu  furnishing  a  simple  approx- 
imation to  what  occurs  in  nature  (see  also  p.  26). 

12 


CONCURRENT  FORCES 


13 


If  a  number  of  vectors  K^,  V^,  •  •  -,  V„,  are  added  by  putting  the 

initial  point  of  each  upon  the  terminal  point  of  the  preceding  (cf. 

Ex.  3,  §  5),  a  polygonal  line  is  formed  which  is  called  the  vector 

polygon,  or  in  particular,  if  the  vectors  are  forces,  the  force  polygon. 

The  resultant  force  is  the  one  which,  starting  from  the  initial  point 

of  the  first  force,  closes  the  polygon. 

Ex.  Show  that  if  we  add  the  n  vectors  Fj,  V^,  ■  •  •>  F„  in  various  orders, 
we  obtain  ;«!  force  polygons,  all  of  which  have  the  same  closing  side. 

Thus  we  may  state :  A  hody  can  he  at  rest  under  the  action  of 
concurrent  forces  when,  and  only  when,  the  force  polygon  is  closed. 
For  then,  and  then  only,  will  the  resultant  be  zero.  Geometrically, 
tliis  is  the  method  usually  used  in  treating  problems  of  equilib- 
rium. The  analytic  statement  of  the  condition  is  found  by  pro- 
jecting the  resultant  upon  the  axes  and  referring  to  Ex.  10,  §  5.  If 
X,  Y,  and  Z  denote  the  projections  of  the  force  F  upon  the  axes, 
the  conditions  for  equilibrium  of  concurrent  forces  takes  the  form 


5)x,=  o,       Xr,=  o, 


1  =  1 


(1) 


in.   PROBLEMS  ON  EQUILIBRIUM  OF  CONCURRENT  FORCES 

[Accompany  each  problem  with  a  diagram.  Each  problem  should  be 
solved  geometrically  and  analytically  when  possible.] 

1.  Show  that  the  forces  represented  by  the  medians  of  a  triangle  are  in 
equilibrium,  the  jxjsitive  sense  of  each  median  being  toward  the  vertex. 
The  solution  follows  as  an  example  of  the  method  of  attacking  these 
problems. 

(a)  Geometric  proof.  As 
the  medians  meet  in  a  point, 
the  f(jrces  are  concurrent. 
We  shall  show  that  the  force 
polygon,  in  this  case  a  tri- 
angle, is  closed.  To  do  this 
we  shift  the  force  C'C  par- 
allel to  itself,  so  that  its 
initial  point   C   falls   upon 

the  terminal  point  B  of  B'B.  Then  C'BDC  is  a  parallelogram  and  CD  = 
C'B  =  AC,  C  being  the  mid-point  of  AB.  To  show  the  force  polygon  is 
closed  we  merely  need  to  show  that  DB'  is  i)arallel  with  and  equal  to  A' A. 
This  will  follow  if  we  can  show  the  two  parallelograms  ACA'B'  and 


'^D 


Fig.  G 


14  STATICS 

RA'DC,  of  which  A' A  and  DB'  are  corresponding  diagonals,  to  be  equal. 
But  CD  is  equal  to  and  parallel  with  AC",  and  so  is  B'A',  for  the  segment 
joining  the  mid-points  of  two  sides  of  a  triangle  is  parallel  with  the  third 
side  and  equal  to  one  half  of  it.  Finally  AB'  —  B'C,  and  hence  the  paral- 
lelograms are  equal  and  the  theorem  is  proven. 

(b)  Analytic  proof.  If  the  coordinates  of  ^,  J5,  and  C  are  respectively 
(xj,  j/i),  (Xj,  y.2),  (Xg,  yg),  the  coordinates  of  A'  are  1  (Xg  +  Xg),  l(y^  +  y^), 
of  B',  \{x^  +  Xi),  1(^3  +  ?/i),  of  C,  1  (xj  +  X2),  ^(z/i  +  2/2)-  The  conii)onents 
or  projections  of  the  forces  A' A,  B'B,  and  C'C  on  the  x-axis  are  therefore 
X^  =  xj  -  1  (X2  +  Xg) ,  X2  =  X,  -  ^  (xg  +  Xi) ,  Xg  =  Xg  -  ^  (Xj  +  Xj).  Heucc 
A'l  +  A2  +  Ag  =  Xj  -I-  X2  -f-  Xg  -  1  (2  Xi  +  2  X2  +  2  Xg)  =  0.  Similarly,  Y^  +  1\ 
+  Fg  =  0,  and  the  forces,  having  a  vanishing  resultant,  are  in  equilibrium. 
In  this  problem  the  analytic  treatment  is  briefer  and  more  elegant,  though 
this  is  by  no  means  always  the  case. 

2.  Prove  that  three  forces  of  magnitude  5,  6,  and  12  can  never  be  in 
equilibrium. 

3.  Let  AB  and  CD  be  diameters  of  a  circle.  Three  concurrent  forces 
are  represented  in  magnitude  and  direction  by  AB,  DC,  and  2BD.  Show 
they  are  in  equilibrium. 

4.  Three  concurrent  forces  are  determined  by  their  magnitudes  and  the 
angles  their  directions  make  with  the  axes  as  follows : 

F^;    /=20,     a  =  30°,     j3  =  G0°,     y  =  90°. 
F^;    f=  14,     a  =  90°,     /3  =  45°,     y  =  45°. 
Fg-,    /=  10,     a:  =  45°,     13  =  45°,     y  =  90°. 
Find  the  force  which  will  hold  them,  in  equilibrium. 

5.  Find  the  resultant  of  the  forces  given  by  their  projections  upon  the 

*^^=  ^1;    /x=5,    /,  =  7,        /,  =  11. 

F^;    /x  =  6,    /,  =  -5,    ./;  =  2. 

n;  /x  =  9,  /.  =  -2,  /.  =  -o. 

G.  Three  concurrent  forces  are  in  equilibrium.  Show  they  lie  in  a  i>lane. 
Show  further  that  if  their  magnitudes  are/j./^,  /g,  and  the  angles  between 
them  are  <T2g,  a^^,  and  o-jg,  then 

/^      =    /"'     =    y^     .  (Lamp's  Theorem) 

smo-og      sino-gj      sinaj2  - 

Hint.  Consider  the  force  triangle. 

7.  Let  a  system  of  forces  be  represented  by  OA ,  OB,  OC,  •  ■  ■,  ON.  Show 
that  if  they  are  in  equilibrium,  the  coordinates  of  O  are  the  averages  of  the 
corresponding  coordinates  oi  A,  B,  C,  ■  • -,  N ;  \i\  other  words,  O  is  the 
center  of  mass  of  a  system  of  equal  jiarticles  at  the  points  A,  B,  C,  ■  •  •,  N 
(see  p.  27). 

8.  A  weight  W  upon  an  inclined  plane  whose  angle  with  the  horizon  is  i, 
may  be  held  at  rest  either  by  a  force  Q  acting  up  the  incline,  or  by  a  force 
P  acting  horizontally.  Show  that  W  =  PQ/VP-+  QS  and  cos  t  =  Q/P. 
Is  your  result  in  poundals  or  pounds  ? 


CON-CUREENT  FORCES  15 

9.  Find  the  angles  between  three  forces  2P,  ZP,  and  4P  which  are  in 
equilibrium . 

10.  Two  rafters  meeting  at  an  angle  60°  in  a  vertical  plane  support  a 
chandelier  weighing  90  lb.    What  is  the  force  along  each  rafter? 

11.  A  body  of  mass  140  lb.  is  attached  to  the  ends  of  two  ropes  of 
length  6  and  8  ft.  respectively,  which  are  fastened  to  a  horizontal  beam  at 
two  points  10  ft.  apart.    Find  the  tensions  on  the  ropes. 

12.  A  bar  weighing  100  lb.  is  suspended  by  chains  passing  from  a  ring 
to  its  ends.  The  chains  make  an  angle  of  45°  with  the  vertical.  Find  the 
tension  on  the  chains. 

Hint.  Consider  the  bar  replaced  by  two  weights  of  50  lb.  each  at  its 
end  ixjints  and  connected  by  a  weightless  rigid  bar. 

13.  A  body  is  kept  at  rest  on  a  smooth  inclined  plane  by  two  forces, 
each  equal  to  half  the  weight  of  the  body,  the  one  acting  horizontally  and 
the  other  directly  up  the  plane.    Find  the  angle  of  inclination  of  the  j)lane. 

14.  Two  smooth  rectangles  have  a  horizontal  edge  in  common,  their 
faces  being  inclined  to  a  horizontal  plane  at  angles  Oj  and  a^  respectively. 
Weights  u\  and  ic^  rest  one  on  each  plane,  and  are  connected  by  a  string 
running  smoothly  over  the  common  horizontal  edge.  If  the  system  is  in 
equilibrium,  find  the  ratio  of  il\  to  w^. 

Hint.  Consider,  say,  the  weight  w^,  remembering  that  the  string  trans- 
mits the  force  exerted  upon  ic^  by  tc^  in  line  with  the  plane  on  which  w^ 
rests. 

Friction.  When  two  bodies  are  in  contact  along  plane  surfaces,  there  is 
usually  a  pressure  keeping  them  togeiher  and  causing  more  or  less  of  an 
interplay  of  the  slight  roughness  of  their  surfaces.  There  thus  arises  a 
certain  resistance  to  any  force  tending  to  make  the  surfaces  slide  over  one 
another.  This  resistance  is  a  force  exerted  by  the  protruding  particles 
against  each  other,  and  has  a  direction  oi:)posite  to  that  of  the  attempted 
motion  and  is  called  a  frictional  force.  As  long  as  there  is  no  motion,  the 
frictional  force  exactly  balances  the  force  tending  to  produce  motion.  But 
it  is  found  that  when  the  moving  force  exceeds  a  certain  limit,  the  body 
does  move,  although  still  retarded  by  the  frictional  force.  Experiments 
justify  as  an  approximation  the  assumption  that  this  force  is  proportional 
to  the  force  pressing  the  two  bodies  together,  that  is,  to  the  normal  com- 
ponent n  of  the  resultant  of  the  forces  acting  at  the  common  surface.  We 
have  then  the  result 

(1)  The  frictional  force  has  a  direction  opjwsite  to  that  of  the  resultant 
of  the  other  forces. 

(2)  It  has  a  magnitude  y  which, 

(a)  when  no  motion  takes  place,  is  equal  to  the  magnitude  of  this 
resultant ; 

(b)  when  motion  takes  place,  is  proportional  to  the  magnitude  of  the 
normal  component  of  their  resultant.     In  symbols  y= /x,n,  where  /*  is  a 


16  STATICS 

proportionality  factor,  and  is  called  the  coefficient  of  friction.  It  depends 
upon  the  character  of  the  surfaces  in  contact.  The  student  should  avoid 
feeling  that  the  above  is  more  than  an  approximation,  and  apply  it  only 
in  the  case  of  bodies  sliding  on  one  another.  There  are  many  other 
kinds  of  frictional  forces,  too  numerous  and  varied  to  receive  attention 
here. 

15.  Determine  the  frictional  force  "which  keeps  a  body  of  weight  1  lb. 
at  rest  on  a  plane  inclined  at  an  angle  of  30°,  the  slipping  point  being  not 
yet  reached. 

16.  If  a  plane  be  tilted  farther  and  farther,  the  normal  component  of 
the  gravitational  force  acting  upon  a  body  on  the  surface  of  the  plane 
diminishes,  while  the  component  tending  to  produce  slipping  increases. 
The  angle  of  inclination  e  at  which  the  slipping  begins,  is  called  the  angle 
of  friction.    Prove  fi  =  tan  e. 

17.  Suppose  a  body  rest  on  a  table  and  a  force  of  magnitude  h  tends  to 
move  it.  The  magnitudes  of  the  friction y  depend  upon  h.  Draw  the  graph 
of  /  as  a  function  of  h  up  to  and  beyond  the  slipping  point.  What  is  the 
value  of  h  at  the  slipping  point  ? 

18.  Let  a  be  the  angle  of  inclination  of  a  plane,  and  f  the  frictional 
force  tending  to  prevent  or  retard  a  body's  sliding  along  the  plane.  Draw 
the  graph  of  /  as  a  function  of  a  up  to  and  beyond  the  angle  of  friction. 

Take  e,  say  at  20°. 

19.  A  mass  of  10  lb.  rests  upon  a  table  and  can  be  just  moved  by  a 
force  of  3  lb.  acting  horizontally.  Find  the  coefficient  of  friction  and  the 
direction  and  magnitude  of  the  resultant  reaction  of  the  plane. 

20.  A  body  of  weight  rv  rests  on  a  rough  horizontal  table.  If  a  force  be 
applied  with  an  upward  component,  this  upward  component  will  lessen  the 
normal  reaction  and  hence  the  friction.  Show  that  the  least  force  which 
will  move  the  body  makes  an  angle  e  with  the  horizontal  and  has  a  magni- 
tude wsine. 

21.  Two  weights  of  10  and  20  lb.  lie  upon  a  rough  inclined  plane  con- 
nected by  a  string  which  passes  around  a  pulley  in  the  plane.  Find  the 
greatest  inclination  of  the  plane  consistent  with  equilibrium  of  the  system, 
the  coefficient  of  friction  being  /x. 

22.  How  high  can  a  particle  rest  in  a  hemispherical  basin  of  radius  r, 
the  coefficient  of  friction  being  fi  ? 

9.  Nonconcurrent  forces;  moments.  We  now  turn  to  the 
consideration  of  a  system  of  forces  acting  at  various  points  of  a 
rigid  body  whose  lines  of  action  do  not  all  intersect  in  one  point. 
Besides  a  tendency  to  translate  there  will,  in  general,  be  a  tendency 
to  rotate  the  body  about  some  line  in  it.  If  there  is  to  be  equilib- 
rium, the  tendency  to  rotate  about  any  line  whatever  in  the  body 


MOMENTS 


17 


must  be  zero.  We  consider  the  question  of  measuring  this  tend- 
ency to  rotate. 

Let  us  first  examine  a  force  whose  line  of  action  is  perpendic- 
ular to  the  axis  of  rotation. 

The  angle  between  two  nonintersecting  lines  is  defined  to  be  the  angle 
between  two  intersecting  lines  which  are  parallel,  one  with  each  of  the 
given  lines. 

The  tendency  of  such  a  force  to  turn  about  the  axis  is  found  to 
be  proportional  to  the  magnitude  /  of  the  force  and  to  the  dis- 
tance p  of  its  line  of  action 
from  the  axis,  or  the  arm  of 
the  force  as  it  is  called ;  in 
other  words,  the  tendency  is 
proportional  to  f  •  ]p. 

The  student  will  find  the  best 
substantiation  of  this  empirical 
fact  in  the  well-known  properties 
of  levers  and  balances.  (See  Mach, 
The  Science  of  Mechanics,  trans- 
lated by  McCormack,  Chicago, 
1902.) 

This  quantity  /  •  ^  is  called 
the  moment  of  the  force  about 

the  axis.  If  a  number  of  forces  are  acting,  some  tending  to  turn  in 
one  direction  and  others  m  the  opposite,  we  fix  on  a  positive  direc- 
tion and  give  a  positive  sign  to  the  moments  of  the  forces  tend- 
ing to  turn  in  that  direction,  and  a  negative  sign  to  the  moments 
of  those  tending  to  turn  in  the  opposite  direction. 

Should  the  force  F  not  be  perpendicular  to  the  axis,  we  resolve 
it  into  two  components,  —  F^  parallel  with  the  axis,  and  F^  per- 
pendicular to  the  axis  (see  Fig.  8). 

The  component  F^  evidently  has  no  tendency  to  turn  about  a 
line  parallel  to  it,  so  that  the  turning  effect  of  F  is  that  of  F^- 
We  therefore  define  the  moment  in  this  case  to  be  f,,-p,  where 
/„  is  the  magnitude  of  F^  and  p  the  distance  of  its  line  from 
the  axis. 


18 


STATICS 


Wlien  the  forces  studied  all  lie  in  one  plane  perpendicular  to 

the  axis  the  problem  is  usually  regarded  as  a  plane  proUem,  and 

we  speak  of  the  moments  of  the  forces  about  a  point,  namely,  the 

point  in  which  the  axis 

pierces  the  plane.    The 

following  exercises  are 

important  parts  of  the 

theory  of  moments  and 

should   be    thoroughly 

studied. 

Ex.  1.  Let  P  be  any 
point  upon  a  force  vector 
For  upon  its  line  of  action. 
Prove  that  the  moment  of  F 
about  a  point  O  is  OP  times 
the  projection  fp>  upon  a  line 
~~""-  perjjendicular  to  OP. 

Hint.  Use  the  previous 
definition,  and  similarly  of 
triangles. 

Ex.  2.  Let  OX  be  an 
axis,  OP  a  perjiendicular 
dropped  upon  it  from  any 
point  P  of  a  force  vector 
F  or  of  its  line  of  action, 
and  let  fpf  be  the  projec- 
tion of  i^  on  a  line  perpendicular  to  the  plane  of  OX  and  OP.  Calling 
OP  =  p',  show  that  the  moment  of  F  about  OX  is  p'fp.. 

Ex.  3.  From  the  preceding,  together  with  I  of  §  .5,  prove  that  the  moment 
of  the  resultant  of  several  concurrent  forces  is  the  sum  of  the  moments  of  the  sepa- 
rate forces. 

Hint.  Take  the  common  point  of  the  lines  of  force  for  P. 
Ex.  4.  Prove  that  if  the  moment  of  F  about  OX  vanishes,  F  and  OX  lie 
in  the  same  plane. 

10.  Couples.  If  two  equal  and  opposite  forces  are  concurrent, 
their  effect  is  nil.  If  not  concurrent,  they  are  said  to  form  a 
couple.  A  couple  has  no  tendency  to  translate  the  body  as  a  whole, 
as  may  be  seen  by  putting  the  body  on  a  track  runnmg  in  any 
given  direction.  Equal  forces  will  be  exerted  on  the  body  in  oppo- 
site directions.    The  couple  will,  however,  have  a  tendency  to  rotate 


Fig.  8 


COUPLES 


19 


the  body.    This  tendency  is  measured  by  the  so-called  moment  of 
the  couple,  which  is  simply  the  sum  of  the  moments  of  its  two 

forces  about  a  point  midway  between  their 

lines  of  action. 


F 


F' 


Fig.  9 


^  Ex.  1.   Prove  that  the  moment  of  a  couple  about  all 

points  of  its^plane  w  the  same,  and  equal  to  p  .f  where 
p  is  the  distance  between  the  jmrallel  forces  and  f  is  their 

—    ^^     common  magnitude. 

■f  Ex.2.  Prove  that  the  "moments  of  a  couple  about 

all  axes  which  have  the  same  direction  is  the  same, 
and  that  this  common  value  is  the  moment  of  one  of 
the  forces  about  an  axis  irith  the  given  direction  through 
the  other  force. 

The  point  of  application  of  a  force  F  may 
be  thought  of  as  changed  from  P  to  Q,  pro- 
vided at  the  same  time  a  couple  be  introduced  consisting  of  a  force 
F  at  F  and  a  force  at  Q  equal 
to  P  in  magnitude  but  oppo- 
site in  direction.  For  the  new 
system  is  merely  the  original 
force  with  two  mutually  an- 
nulling forces  introduced.  It 
follows  that  any  system  of 
nonconcurrent  forces  may  be 
replaced  by  a  system  that    ^ 


is  concurrent  at  any 
given  point  0  together 
with  a  system 
of  couples. 
Thus  the  effect 
of  a  system  of 
nonconcurrent 
forces  is  the 
same  as  that  of 


/ 
/ 


-^A 


Fig.  10 


(1)   a  single  force  applied  at  any  given  point  0  and  equal  to  the 
resultant  the  given  forces  ivould  have  if  concurrent,  and 


20  STATICS 

(2)  a  set  of  couples,  one  for  each  force  of  the  given  system,  and 
each  having  for  its  moment  about  any  axis  the  moment  of  the  cor- 
responding force  about  a  line  through  0  parallel  to  the  axis  (see 
Ex.  2). 

The  moments  of  the  couples  about  an  axis  through  0  will,  in 
general,  vary  with  0,  because  to  change  the  point  of  application 
0  of  the  resultant  means  to  introduce  a  new  couple.  Only  in 
case  the  resultant  vanishes  will  this  new  couple  vanish,  and  the 
sum  of  the  moments  of  the  couples  be  independent  of  the  point 
0  chosen. 

11.  Equilibrium.  A  set  of  forces  is  said  to  be  in  equilibrium  if 
a  resting  body  remains  at  rest  under  their  action.  Having  studied 
the  tendencies  of  forces  to  produce  motion,  we  may  state  the  fol- 
lowing necessary  and  sufficient  conditions  for  equilibrium 

(1)  the  resultant  of  all  the  forces,  regarded  as  concurrent, 
vanishes  ; 

(2)  the  sum  of  the  moments  of  the  forces  abotit  any  axis  vanishes. 

Our  next  task  is  to  express  these  conditions  analytically.  If  the 
forces  F^,  F^,  ■  •  •,  F^  be  projected  on  to  the  three  coordinate  axes, 

giving  /i,,  /a^,  •  •  •,  /„x ;  fly,  fly,  •  •  -,  fny  \  Az'  A.'  *  "  ',  fnz,  ^6  havc, 
referring  to  I,  §  5, 

fly  +  fly+---+fny=Q)-,        OX        ^fiy=0 

1 

These  conditions  express  the  demand  that  there  shall  be  no 
tendency  to  translate  the  body  as  a  whole.  If  they  are  satisfied, 
the  forces  may  be  grouped  in  couples  whose  effect  to  rotate  about 
an  axis  remains  the  same,  no  matter  how  the  axis  is  shifted  parallel 
to  itself.  We  may  therefore  take  it  through  the  origin.  If  the 
sum  of  the  moments  of  the  couples,  or,  what  is  the  same  thing,  of 
the  original  force  system  is  to  vanish  when  taken  about  any  axis, 
it  must,  in  particular,  when  taken  about  the  three  coordinate  axes. 


EQUILIBRIUM 


21 


But,  conversely,  if  the  sum  of  the  moments  about  each  coordinate 
axis  vanishes,  it  vanishes  about  every  axis. 
Ex.  Prove  this.    Use  Ex.  4,  §  9. 

Hence  we  have  the  second  condition  in  the  form :  the  moments 
of  the  forces  about  the  three  coordinate  axes  must  vanish.    We 


Fig.  11 

proceed  now  to  calculate  analytically  the  sums  of  the  moments 

about  the  axes.    Let  us  agree  to  call 

a  positive  rotation  about  an  axis  one 

which  would   drive  a  right-handed 

screw  forward  in  the  direction  of  the 

axis,  as  denoted  in  the  accompanying 

figure.    Each  force  is   spht  into  its 

components,  and  the  moments  of  each 

component  about  the  axes  are  tabulated  in  the  above  scheme, 


Axis 

COMPOXENT 

F. 

Fy 

F. 

OX 

0 

-Zfy 

+  yfz 

OY 

+  Z/x 

0 

-x/. 

OZ 

-  yfx 

+  Xfy 

0 

22 


STATICS 


(x,  y,  z)  being  the  coordinates  of  the  point  of  application  of  the 
force. 

We  have,  therefore,  writing  down  these  moments  for  all  the 
forces  and  forming  the  sums,  the  second  set  of  conditions  for 
equilibrium 


(-?i/ix  -  J^i/i.)  +  (^2/2.  -  ^2/2.)  + 


+  {Vnfnz  -  2„f^y)  =  0, 
+  {Znfnx  -  X„f„,)  =  0, 
+  (X^fny  —  Vnfnx)  =  0, 


or 


.  XiZifi.  -  xj,:)  =  o[.  II 


Ex.  1.  Show  that  the  moment  of  a  force  about  a  point  on  its  line  of 
action  vanishes;  also  about  any  axis  intersecting  its  line  of  action  or 
parallel  to  it. 

Ex.  2.  Show  that  the  moment  about  the  origin  of  the  force  F  with 
projections /j  vluA  fy  on  the  axes  and  with  point  of  application  (x,  y^  is 
xfy  —  yfj.  as  follows:  find  the  equation  of  the  line  of  action,  find  its  dis- 
tance from  the  origin,  and  multiply  by  the  magnitude  /  of  the  force. 

Ex.  3.  Show  that  if  three  forces  acting  on  a  rigid  body  are  in  equilib- 
rium (a)  they  lie  in  a  plane  ;  (b)  they  are  either  concurrent  or  parallel ;  and 
(c)  the  force  triangle  is  closed,  the  relation  between  the  forces  being  given 
by  Lamp's  theorem  (cf.  problem  6,  §  8). 

Ex.  4.  The  composition  o/  parallel  forces.  The  resultant  has  been  defined 
only  for  concurrent  forces.  A  natural  extension  of  the  idea  to  parallel 
forces  is :  the  resultant  of  a  set  of  i^arallel  forces  is  the  force  which  would 
hold  the  set  in  equilibrium,  tcith  its  direction  reversed.  By  the  theory  of 
moments  show  that  the  resultant  of  two  parallel  forces  F^  and  F^,  with 
magnitudes  /j  and  f^,  has  a  magnitude  f  +f^  or  \f^  ~f-2\  (^^e  bars  denot- 
ing the  absolute  value  of  the  diiference),  according  as  the  directions  of 
the  forces  are  the  same  or  opposite,  and  that  the  point  of  application  of  the 
resultant  divides  the  line  joining  the  points  of  application  in  the  ratio ^  :/j, 
internally  or  externally,  according  as  the  forces  are  similarly  or  oppositely 
directed. 

Ex.  5.  Generalize  to  three  parallel  forces  not  all  in  the  same  plane. 
Consider  the  points  where  their  lines  cut  a  perpendicular  plane,  and  let 
their  coordinates  be  (x^,  y^),  (x„,  y^),  (x^,  y^).  Find  first  the  point  of 
application  of  the  resultant  of  two  of  the  forces  and  then  bring  in  the 
third. 


EQUILIBEIUM 


23 


IV.   PROBLEMS  ON  MOMENTS  AND  THE  EQUILIBRTOM  OF  FORCES 

[In  solving  problems  in  equilibrium  of  forces  the  student  should  accom- 
pany each  problem  by  a,  carefully  drawn  diagram  in  which  are  marked  all 
forces  acting  on  the  body,  or  on  each  separate  body  in  case  more  than  one  is 
involved,  being  careful  to  include  "  reactions  "  (pressures  or  tensions)  at 
all  points  where  bodies  are  in  contact,  tensions  of  strings,  and  so  forth. 
Then  he  should  apply  the  conditions  for  equilibrium  to  the  body,  or,  in  case 
of  several,  to  each  body  and  to  the  system  as  a  whole.  Geometric  relations 
in  the  diagram  should  also  be  written  down.  The  result  will  be  a  system 
of  equations  to  be  solved  for  the  unknown  forces.  The  number  of  equa- 
tions should  agree  with  the  number  of  unknowns  if  the  problem  is  deter- 
minate. Notice  that  to  a  force  in  a  plane  correspond  two  unknowns,  and 
in  space  three,  though  this  number  is  diminished  when  either  the  magni- 
tude or  direction  or  other  similar  data  is  given  with  respect  to  the  force. 
In  the  following  problems  the  forces  all  lie  in  one  plane,  and  moments 
need  only  be  taken  with  respect  to  one  point.] 

1.  A  uniform  rod  AB  of  length  21  and  weight  w  rests  with  the  end  A 
against  a  smooth  vertical  wall,  while  to  the  lower  end  B  is  fastened  a 
string  BC  of  length  2  i,  coming  from  a  point  C  in  the  wall  directly  above 
A .  If  the  system  is  in  equi- 
librium, determine  the  angle 
ACB^e.  Show  that  the 
tension  on  the  string  T^ 
w  ■  sec  6,  and  that  the  pres- 
sure against  the  wall  P  = 
IV  ■  tan  6. 

Solution.  We  consider  the 
forces  acting  on  the  bar.  A  s 
the  wall  is  "  smooth,"  the 
pressure  between  the  bar 
and  the  wall  is  perpendicu- 
lar to  the  wall,  for  otherwise 
there  would  be  slipjnng. 
The  weight  w  may  be  con- 
sidered as  acting  directly 
downward  at  the  mid-point 
of  the  bar,  and  the  tension 
T  acts  along  the  string. 
Applying  the  first  condi- 
tions of  equilibrium,  the  sum  of  the  horizontal  components  must  vanish: 

P-Tsme=0;  (1) 

and  also  the  sum  of  the  vertical  components : 

-w-{-Tc.o»6  =  0.  (2) 


24  STATICS 

Also,  taking  moments  about  ^, 

-wxlsin<f,-^  Tx2lsm(ABC)  =  0.  (3) 

The  unknowns  are  0,  P,  T,  and  <^  and  (ABC).    We  therefore  need  two 
more  relations.    These  relations  are  geometric  in  character,  and  are 

(ABC)  ^^-6,  (4) 

and,  from  the  law  of  sines,  /^  =  — —  •  (5) 

We  now  proceed  to  answer  the  questions  proposed.  Equation  (2)  solved 
for  T  gives  T  =  w  sec  6,  one  of  the  required  results.  This  value  substituted 
in  (1)  gives  P  =  w  tan  6,  a  second  required  result.  To  determine  0  we  sub- 
stitute the  value  of  T,  and  the  value  of  ABC  given  by  (4)  in  equation  (3), 
at  the  same  time  dividing  by  wl  and  expanding  sin  (<f>  —  6): 
—  sin  (f>  +  2  sec  0  (sin  <f>  cos  6  —  cos  <^  sin  6)  =  0, 
or  —  sin  ^  +  2  sin  (f>  —  2  cos  <f>  tan  ^  =  0,  ^ 

that  is,  tan  <^  =  2  tan  0. 

Equation  (5)  may  be  written     sin  <j>  =  -  sin  6. 

Unless  6=0,  we  may  divide  this  equation  by  the  preceding  and  obtain 

21 


cos  4>  =  -^ —  cos  6. 


Squaring  and  adding,  we  have 


-^(4  sin2^  +  cos^^)  =  -^  (1  +  .3  sin2^). 


.     „             4/-'-i-2 
whence  sinp  =  ±-\/ — •• 

The  excluded  value  ^  =  0  also  gives  a  position  of  equilibrium,  so  that  the 

final  required  result  is 

/JO             a             ■       M ''  -  ^- 
a  =  0,     or     &  =  arcsin-y/ — 

Remarks,  (a)  It  is  not  necessarj^  that  the  comjwnents  of  the  forces  along 
horizonlal  and  vertical  lines  be  taken  in  the  first  conditions  of  equilibrium. 
Any  two  intersecting  lines  will  do.  In  this  problem  horizontal  and  vertical 
directions  were  chosen  because  two  of  the  forces  had  those  directions. 

(b)  Moments  might  have  been  taken  about  any  point.  A  was  chosen 
because  thereby  the  unkno^^•n  force  P  was  eliminated  from  the  equation. 

(c)  In  choosing  the  geometric  relations  we  did,  we  were  guided  by  the 
necessity  of  having  two  more  relations  in  «^  and  (ABC)  in  order  to  elimi- 
nate these  unknowns  from  equation  (3). 

(d)  It  is  interesting  to  ask  whetlijer  the  equilibrium  is  possible.  This  is 
the  case  if  a  real  value  of  6  has  been  obtained,  for  then  all  the  conditions 
are  satisfied.  In  order  that  6  be  real  the  expression  under  the  radical  sign 
must  be  positive  and  less  than  1.    This  will  be  the  case  if  b  lies  between  I 


EQUILIBRIUM  25 

and  2  I;  that  is,  the  string  must  be  longer  than  the  bar  but  less  than  twice 
its  length. 

2.  Weights  of  1,  2,  3,  4,  and  5  lb.  act  on  a  bar  at  distances  1,  2,  3,  4, 
and  5  ft.  from  one  end.  Find  the  magnitude  and  point  of  application  of 
the  force  which  will  hold  the  system  in  equilibrium  (neglecting  the  weight 
of  the  bar). 

3.  A  bar  of  uniform  thickness  weighs  10  lb. and  is  5  ft.  long.  Weights  of 
9  and  5  lb.  are  hung  from  its  extremities.  On  what  point  will  it  balance? 
(Assume  the  weight  of  the  bar  to  act  downward  from  its  center) . 

4.  Find  the  true  weight  of  a  body  which  weighs  8  oz.  and  9  oz.  in  the 
right  and  left  pans  of  a  false  balance.  (The  balance  is  "false  "  because 
the  two  arms  of  the  beam  are  of  unequal  length.) 

5.  A  span  of  a  bridge  40  ft.  long  weighs  10  T.  and  is  supported  by  two 
similar  piers  at  the  ends.  What  is  the  thrust  on  each  pier  if  a  wagon  weigh- 
ing 2  T.  is  (a)  at  the  middle  of  the  span?  (b)  two  thirds  the  way  across? 

6.  A  imiform  bar  2  ft.  8  in.  long  weighing  5^  lb.  is  supported  by  a 
smooth  peg  at  one  end  and  by  a  vertical  string  4  in.  from  the  other  end. 
Find  the  tension  of  the  string,  the  reaction  of  the  peg,  and  the  inclination 
of  the  bar.  (Xote  that  when  anything  is  smooth,  or  frictionless,  the  force 
is  normal  to  the  surfaces  in  contact.) 

7.  A  rope  of  length  I  is  tied  to  a  column  and  a  man  is  pulling  at  the 
other  end.  If  he  exerts  a  force  independent  of  the  direction  of  the  rope, 
at  what  point  of  the  column  should  the  rope  be  attached  in  order  that  the 
man's  efforts  be  most  effective  in  overturning  the  column  ?  (Assume  that 
the  man's  hands  are  on  a  level  with  the  base  of  the  column.) 

8.  A  gate  is  hung  in  the  usual  manner  by  two  hinges  on  a  gate  post. 
Indicate  the  forces  acting  on  the  gate  when  it  hangs  open  and  in  equilib- 
rium, and  show  how  it  may  happen  that  the  reaction  on  one  of  the  hinges 
is  wholly  horizontal. 

9.  A  cylinder  of  length  2  I  and  radius  ?•  rests  with  a  point  of  one  base 
on  a  rough  floor  and  Avith  a  point  of  the  other  base  against  a  smooth  wall, 
so  that  its  axis  lies  in  a  plane  perpendicular  to  the  floor  and  wall  and 
makes  an  angle  i  with  the  floor.  Find  the  frictional  force  on  the  floor 
which  keejis  it  in  this  position,  and  show  that  it  vanishes  when  tan  i  =  l/r. 
What  does  this  result  mean  with  respect  to  the  center  of  tlie  cylinder? 

10.  A  uniform  rod  is  suspended  from  a  hook  by  two  strings  of  length  a 
and  h  tied  to  its  ends.  Show  that  in  a  position  of  equilibrium  the  tensions 
on  the  strings  are  proportional  to  a  and  h. 

11.  A  weightless  rod  AB  of  length  I  can  turn  freely  in  a  vertical  plane 
about  A .  A  weight  w  is  suspended  from  a  ])oint  C  of  the  rod  distant  c 
from  A.  A  string  attached  at  B  and  making  an  angle  150°  with  the  rod 
holds  it  in  equilibrium  in  a  horizontal  position.  Find  the  tension  on  the 
string  and  the  magnitude  of  the  pressure  at  A. 


26  STATICS 

12.  A  uniform  thin  rough  rod  passes  under  one  peg  and  over  a  second 
higher  one,  its  center  being  above  the  higher  peg  and  distant  from  the 
higher  and  lower  pegs  a  and  h  respectively.  Show  that  if  the  rod  is  upon 
the  point  of  motion,  the  coefficient  of  friction  is  fx.  =  (h  —  a) tan i/(b  +  a), 
where  i  is  the  inclination  of  the  rod. 

For  further  problems  in  equilibrium  the  student  is  recommended  to 
study  the  theory  of  balances  and  scales,  such  as  freight  and  postal  scales, 
also  derricks  and  hoisting  devices.  Examples  of  the  kind  are  not  given 
here  because  of  the  space  necessary  to  describe  each  apparatus.  Moreover, 
it  will  be  of  great  value  to  the  student  to  formulate  his  own  problem. 

12.  Centers  of  mass  of  systems  of  particles.  We  shall  be  con- 
cerned with  a  system  of  particles  connected  by  weightless  bars. 
A  particle  is  the  idea  derived  by  imagining  all  the  mass  of  a  body 
concentrated  at  a  point,  whereas  a  weightless  bar  is  to  be  thought 
of  as  a  device  for  keeping  the  distance  between  two  particles  con- 
stant. Whereas  such  things  do  not  exist  in  nature,  problems  are 
simplified  by  their  use,  and  the  results  obtained,  if  not  strictly 
faithful  to  actuality,  are  frequently  in  error  by  less  than  the  errors 
of  observation ;  *  moreover,  they  furnish  a  basis  for  more  rigorous 
developments.  An  example  of  the  latter  use  of  such  notions 
appears  in  the  next  section. 

Near  the  earth's  surface  the  attraction  of  gravity  furnishes  an 
example  of  a  set  of  parallel  forces  acting  on  the  above-mentioned 
system  with  magnitudes  proportional  to  the  masses.  We  ask.  What 
force  will  hold  this  set  in  equilibrium  ?  Let  the  particles  have 
masses  m^  m„,  •  •  •,  m„,  and  be  situated  at  points  (x^,  y^,  z^), 
(x^,  2/2,  z.^),  ■  •  -,  («„,  ?/„,  zj.  Call  the  balancing  force  F,  with  pro- 
jections f^,  fy,  f^  upon  the  coordinate  axes,  and  with  point  of 
application  (i,  y,  z).  Applying  the  conditions  I  of  §  11,  in  which 
the  balancing  force  is  included,  we  have 


U 


*  As  an  illustration  consider  a  symmetrical  fly  wlieel  witli  its  cylindrical  axle, 
which  Ave  sliall  suppose  to  protnxde  more  on  one  side  than  on  the  other.  The  gravita- 
tional forces  acting  on  the  system  may  be  considered  as  acting  on  a  particle  with  the 
mass  of  the  wheel  situated  at  its  geometric  center  and  connected  by  a  weightless  bar 
to  a  particle  with  tlie  weight  of  the  axle  situated  at  its  geometric  center. 


CENTERS  OF  MASS  27 

where  M  is  the  total  mass  of  the  system.  This  gives  F  by  its 
components.  Its  point  of  application  is  involved  in  conditions  II, 
into  which  we  introduce  the  projections  of  F  already  determined : 

[y(.%)-  0]  +  [y,(-m,^)-  0]  +  [2/,(-m,5r)_  0]+  .  •  •  =  0, 
[0  -  X  (.%)]  +  [0  -  ^,  (-  m^g)]  +  [0  -  a;,  (-  m^g)]  +  •  •  •  =  0, 
[0-0]  +[0-0]  +[0-0]  +---  =  0, 

_      99,yOn,      Vy,m..  _      Va.\m.- 

or  y  =  -^^ =  =^ J     and     x  =  ^^ • 

-^  Mg  M  M 

Thus  the  point  of  application  is  not  determined,  for  the  conditions 
of  equilibrium  will  be  satisfied  for  any  z.  The  explanation  is  that 
the  point  of  application  of  the  balancing  force  may  be  shifted  to 
any  point  of  its  line  of  action.  Suppose  we  ask  whether  there  is  a 
point  at  which  the  body  will  balance  even  though  turned  in  a 
different  direction,  or,  what  amounts  to  the  same  thing,  when  the 
parallel  forces  on  the  particles  act  in  a  different  direction,  say  oppo- 
site to  that  of  the  ^-axis.    We  should  then  have  for  each  mass 

and  conditions  I  would  give  us 

f.=  ^ff>    /.=  0,    f  =  0; 
while  conditions  II  would  yield 

it  is  now  X  undetermined.    If,  however,  we  take  the  point 


X^'^'    -_X^     __x^ 


{M=Xm. 


M  ^  M  M 

the  body  will  balance  at  this  point  against  a  system  of  forces  par- 
allel to  either  of  the  two  axes  considered,  and,  as  may  easily  be 
proven,  against  any  set  of  parallel  forces  with  magnitudes  vary- 
ing as  the  masses.  Tliis  point  is  called  the  center  of  mass.  (The 
expressions  centroid  and  center  of  gravity  are  also  frequently  used, 
but  sometimes  with  slightly  different  meanings,  so  that  we  shall 
confine  ourselves  to  the  term  given.)    The  center  of  mass  is  of 


28  STATICS 

great  importance  in  the  mechanics  of  systems  of  particles  and  of 
bodies. 

Ex.  Prove  the  assertion  just  made,  that  the  system  will  balance  at  its 
center  of  mass  for  any  direction  of  the  parallel  forces.  Let  the  direction 
cosines  of  the  direction  of  the  forces  be  cos  a,  cosyS,  and  cosy.  Conditions 
I  give  the  comj)onents  f^,  f^^,  f^  of  the  balancing  force,  and  in  conditions  II 
it  will  be  found  that  if  the  values  given  above  for  x,  y,  z  are  used,  the 
coefficients  of  cos  a,  cos  ^,  and  cos  y  vanish  separately,  so  that  the  conditions 
are  fulfilled  no  matter  what  these  cosines  be.  Show  that  the  magnitude  of 
the  balancing  force  is  Mg. 

V.   PROBLEMS  ON  THE  CENTER  OF  MASS 

1.  A  ball  of  2  lb.  and  one  of  20  lb.  are  fixed  to  the  ends  of  a  uniform 
bar  5  ft.  long  and  of  weight  5  lb.  Find  the  center  of  mass  of  the 
system. 

2.  A  fly  wheel  weighing  2  T.  rests  upon  an  axle  of  400  lb.,  3  ft.  long, 
the  plane  of  the  wheel  dividing  the  axle  into  two  parts  in  the  ratio  of  1 
to  3.    Find  the  center  of  mass  of  the  system. 

3.  Given  two  masses  wij  and  m^  situated  at  the  points  (xj,  y^,  Zj)  and 
(Xg,  y^,  Zj)  respectively,  show  by  considering  moments,  that  the  center  of 
mass  divides  the  segment  joining  the  masses  in  the  ratio  of  m^  to  jBj. 
Hence,  applying  a  formula  of  Analytic  Geometry,  obtain  the  formulas  for 
X,  y,  z  for  two  masses.  A  third  mass  may  now  be  added  by  considering  the 
first  two  united  in  their  center  of  mass,  and  applying  the  same  method. 
Generalizing  thus,  obtain  an  independent  proof  of  the  formulas  for  x,  y,  z. 

4.  Show  that  the  center  of  mass  of  three  equal  particles  at  the  vertices 
of  a  triangle  is  at  the  intersection  of  the  medians. 

5.  The  vertices  of  a  square  carry  weights  1,  1,  1,  and  2.  Determine  the 
position  of  the  center  of  mass. 

13.  Centers  of  mass  of  continuous  bodies.  If  equal  volumes 
taken  from  all  parts  of  a  body  weigh  the  same,  the  body  is  said  to 
have  uniform  density,  and  the  density  is  defined  to  be  the  constant 
ratio  of  mass  to  volume.  If,  however,  there  is  no  such  constant 
ratio,  we  have  recourse  to  the  method  of  derivatives,  saying  first, 
the  average  density  of  a  portion  of  a  body  is  the  mass  of  that  por- 
tion divided  by  its  volume;  and  secondly,  the  density  at  a  point  is 
the  hmit  of  the  average  density  of  a  volume,  including  the  point, 
as  the  dimensions  of  the  including  volume  approach  zero.  Thus 
for  a  nonuniform  body  the  density  will  usually  vary  from  point 


CONTINUOUS  BODIES  29 

to  point,  that  is,  will  be  a  function  of  x,  y,  and  z,  for  which  we 
shall  use  the  notation  d  =  h  {x,  y,  z). 

Turning  now  to  the  problem  of  determiniag  the  centers  of  mass 
of  continuous  bodies,  we  imagine  the  body  split  up  into  cuboids  by 
planes  parallel  with  the  coordinate  planes.  The  mass  of  one  of  the 
cuboids  will  be  approximately  the  volume  multiplied  by  the  den- 
sity 8  {x,  y,  z),  say  at  the  mid-point  of  the  cuboid.  We  have,  there- 
fore, for  the  a:;-coordinate  of  the  center  of  mass  of  the  system  of 
cuboids,  according  to  the  preceding  paragraph,  approximately 

x  = 


h{x,  y,  z)AxAyAz 

the  summations  being  extended  over  all  cuboids  containing  masses, 
and  no  others.  We  obtain  now  the  center  of  mass  of  the  given 
continuous  body  by  passing  to  the  limit 


^x'sJ^^^AxA^jAz      fff''^^'''  ^'  ^)  ^'^''^''^ 


M  M 

and  similarly 

31  ~  M  ' 

hm^^g.g(a^  y,-z)AxAyAz  ^  fff'^^'"'  y,^)dxdydz 
M  "  M 


y  = 


z  = 


where 

M  =  lim ]^ ^ ]^ 8 {x,  y,  z)  A.xAyAz  =   j  j  j  8{x,y,z) dxdydz 

is  the  total  mass  of  the  hody.  In  all  the  integrals  the  limits  of 
integration  are  determined  by  the  surfaces  bounding  the  body,  just 
as  in  the  volume  problems  of  the  Integral  Calculus. 

Ex.  Show  that  the  formulas  of  §§  12  and  13  hold  also  for  oblique  axes, 
except  that  M  no  longer  admits  the  interpretation  of  being  the  mass. 


30  STATICS 

VI.   PROBLEMS  ON  CENTERS  OF  MASS  OF  CONTINUOUS  BODIES 

Unless  the  contrary  is  specified,  the  body  is  to  be  assumed  homogeneous. 

1.  Find  the  center  of  mass  of  a  hemisphere  of  radius  r  whose  density 
varies  as  the  distance  from  the  plane  surface. 

Solution.  Choosing  the  plane  as  xy-plane  with  origin  at  the  center  of  the 
sphere,  we  see  at  once  by  symmetry  that  x  =  y  —  0,  whereas  by  the  above 
formulas,  since  Z(x,  y,  z)  =  kz, 

M  J-r    J-Vr2-aJ       «/ 0 

X+r    /i  +  -\/r2  — at*    n -Jrt  —  jft  —  jfl 
I      I  zdzdydx, 

whence,  as  the  student  may  verify  by  evaluating  these  integrals, 

,,      kirr*'  J  _      kirr^      kirr*      8r 

M  = and         z  = ; =  - — 

8  15  8  15 

2.  Determine  the  center  of  mass  of  a  homogeneous  tetrahedron. 

Hint.  Oblique  axes  should  be  used.    The  bounding  planes  may  then  be 
written  x  =  0,  y  =  0,  z  =  0,  x/a  +  y/b  +  z/c  —  1. 
y    3.  Find  the  center  of  mass  of  a  homogeneous  hemisphere. 

4.  Find  the  center  of  mass  of  that  half  of  the  ellipsoid  x^/a-  +  y-/lP-  + 
-I.  I  (A  —  \  w'hich  lies  to  one  side  of  the  plane  z  =  0. 

5.  Find  the  center  of  mass  of  that  part  of  the  paraboloid  x- /ifi  +  y'^/h'^ 
=  2  z/c  which  lies  below  z  =  h. 

6.  Find  the  center  of  mass  of  a  plate  in  the  form  of  a  sector  of  a  circle 
of  radius  r  and  angle  2  a.  In  particular,  put  a  =  7r/2  and  get  the  centroid 
of  a  semicircular  plate.  For  a  =  tt  the  centroid  should  lie  at  the  geometric 
center.    Check  the  result  in  this  way. 

14.  Centers  of  mass  of  some  bodies  of  special  shapes.  The 
formulas  of  the  preceding  sections  take  on  simpler  forms  in  special 
cases.  Some  of  these  we  here  mention,  leaving  most  of  the  devel- 
opments to  the  student. 

(a)  Cylinders,  or  bodies  bounded  by  a  cylindrical  surface  and 
two  parallel  planes,  the  density  being  the  same  along  all  lines 
parallel  to  the  elements  of  the  cylindrical  surface.  The  last  means 
that  the  density  depends,  if  we  take  the  2;-axis  in  the  direction  of 
the  elements,,  on  x  and  y  only,  i.e.  d  =  8  (x,  y).  If  the  ^tz-plane  be 
taken  halfway  between  the  parallel  bases,  we  see,  by  symmetry, 
that  2;  =  0,  and  furthermore  we  may  carry  out  one  integration  in 
the  other  formulas : 


CENTERS  OF  MASS  31 

h 

I  I  I     x8  (x,  y)  dzdacdy 


X  = 


M 


7i  j  j  x8  (x,  y)  dxdy 


M 


and  similarly, 


M  =  h  j  i  8(x,  y)  dxdy, 
hi  j  y8(x,y)dxdy 


M 


The  form  of  the  integrals  suggests  "  the  center  of  mass  of  an  area," 
and  this  expression  is  sometimes  used,  though  its  real  meaning 
should  be  kept  in  mind.  The  quantity  li  ■  8  (x,  y)  is  sometimes 
called  the  surface  density.  Low  cylinders  are  occasionally  called 
plates. 

(b)  Straight  wires.  Take  the  a."-axis  through  the  wire.  Then  8  (x) 
means  the  "  linear  density  "  (that  is,  the  mass  of  a  piece  of  the 
wire  divided  by  its  length,  or,  in  case  this  quotient  varies,  its  limit). 
The  student  should  show  that 

lim  ^x8  {x)Ax      j  '^^  (•^■)  d^ 

.    ~  lmi^8{x)Ax  ^ 

where  31=  j  8{x)dx. 

(c)  Bodies  of  revolution,  in  which  the  density  is  the  same  at  all 
points  of  any  plane  perpendicular  to  the  axis  of  revolution.  These 
are  a  simple  generalization  of  the  preceding,  in  which  8{x)  means 
the  mass  of  a  plate  bounded  by  two  planes  perpendicular  to  the 
axis  and  a  distance  Ax  apart,  divided  by  Ax,  or,  in  case  this  quan- 
tity varies,  8  {x)  is  its  limit  as  Ax  —  0.  In  particular,  if  the  volume 
density  is  constant,  say  k,  then  8(x)  is  k  times  the  area  of  a  cross 
section  a  distance  x  from  the  origin.  The  same  formulas  of  course 
hold. 


32 


STATICS 


Example.  Consider  problem  1  of  the  preceding  paragraph.  Taking 
the  X-axis  as  the  axis  of  revolution  of  the  hemisphere,  we  have  8(a;)  =  den- 
sity times  the  area  of  a  cross  section  =  kx(yr^  —  x^)^.    Whence 


k  i   x^  (r-  —  X-)  dx      Q 
k  rx(r^-x^)dx 


r 
15" 


The  student  will  notice  how  much  more  simply  the  integrals  are  evaluated 
by  this  method  whenever  it  applies. 

(d)  Bodies  with  an  axis  such  that  there  is  a  set  of  parallel  planes 
which  cut  from  the  body  plates  whose  centers  of  mass  lie  upon  the 
axis.    Taking  the  axis  for  ic-axis,  and  the  other  axes  in  a  plane 


Fig.  13 

parallel  with  the  set  of  planes  mentioned  (the  axes  may  be  oblique ; 

compare  the  exercise  at  the  end  of  §  13),  these  bodies  admit  of  a 

treatment  like  wires.     Then   8(x)  means  the  mass  of  the  body 

between  two  of  the  planes  of  the  set,  corresponding  to  x  and  to 

X  +  Ax,  divided  by  Ax,  or,  if  this  ratio  varies,  its  limit.    The  same 

formulas  again  hold. 

Example.  Any  homogeneous  cone,  right  or  oblique,  has  its  center  of  mass 
J  the  distance  from  its  vertex  to  its  base.    Planes  parallel  with  the  base  make 


CENTERS  OF  MASS  33 

similar  sections  whose  areas  vary  with  the  square  of  the  distance  from  the 
vertex.    We  have,  therefore,  8(x)  =  kx^,  whence 


k( : 


x^dx      o 


k  I  x^dx 
Jo 

(e)  Surfaces  of  revolution.  Since  a  surface  has  no  thickness,  mate- 
rial surfaces  are  idealizations,  though  they  are  nearly  realizable  in 
the  case  of  bodies  made  of  thin  metal.  We  shall  suppose  the  mate- 
rial of  constant  thickness  and  density.  Then,  the  x-axis  being  chosen 
to  coincide  with  the  axis  of  revolution,  and  the  meridian  curve 
being  given  by  y  =f{x:),  we  have  Ac  =  Va«"  4-  Ay^  for  a  chord  of 
the  meridian  curve,  2  7r(y  +  Ay/2)  Ac  for  the  surface  formed  by 
revolving  the  chord  about  the  axis  (see  the  section  on  surfaces  of 
bodies  of  revolution  in  the  Integral  Calculus),  so  that  if  d  is  the 
surface  density,  2  7r(i/ +Ai//2)^l  +  {Ai//Ax)'^Ax  ■  d  is  the  mass 
of  such  a  surface.    Thence 


>^ 


^\dx 
dx, 


and,  by  symmetry,  y  =  0,  s  =  0. 

Compare  this  with  the  derivation  of  the  formula  in  calculus  for  the  area 
of  a  surface  of  revolution.  It  need  scarcely  be  pointed  out  that  the  dis- 
tinction between  this  case  and  case  (c)  is  that  here  a  shell  is  meant,  whereas 
in  case  (c)  the  solid  hounded  by  such  a  shell  was  intended. 

(f)  Curved  wires.  The  equations  of  the  curve  which  gi^•e  the 
form  of  the  wire  should  be  put  into  parameter  form,  say,  x  =f{t), 
y  =  g{t),  z  =  h{t).    Tlien  we  have  for  a  chord 


^c=V(^^fH^yf+(M'=^^^ 


^m\i^\\t. 


At)      \Atj 


34  STATICS 

Heuce  the  mass  of  a  short  piece  of  the  arc  is 

We  therefore  have,  remembering  that  lim  As /Ac  =  1, 


X  = 


\ir.^^,)^s        ^.^ 


M 


J  fit)  8  (0  V/'^  (t)  +  f'  (t)  +  h''  (t)  dt 


M 


and  similarly,       y  = , 

Jh  (t)  5  (t)y/f'^t)  +  g'\t)  +  h'\t)dt 

z  = ■ ' 

M 


where  M  =JB(t)  ^/(gVMV^ )  dl 


\dt/      \dt 


If  possible,  the  length  of  the  cur^'e  should  be  taken  as  parameter, 
for  then  the  radical  reduces  to 


i 


^•y+  /^V+  /^V=  '^^ 


ds/      \ds/      \dsj      ds 


(g)  Bodies  ivhose  bounding  surfaces  are  simpler  in  some  special 
coordinate  system  may  he  treated  hy  the  use  of  such  a  system.  We 
illustrate  by  a  plate  bounded  by  a  cylindric  surface  whose  generating 


CENTERS  OF  MASS  35 

curve  is  given  in  polar  coordinates  (see  the  section  on  areas  bounded 
by  curves  given  in  polar  coordinates  in  the  Integral  Calculus). 
The  formulas  become  (cf.  case  (a)) 


ff  p  cos  e8(p,  6) pdpdd 

^~  M  '' 

CC p  sin  0  8  (p,  6)  pdpd0 

y  =  — i 

i=0, 


where  M^        S(p,e)pdpde. 


JJs(p,e)pdpc 


Example.  Find  the  center  of  mass  of  a  plate   of    uniform  density 
bounded  by  the  cardioid  p  =  a(l  —  cos^).    We  have 

/»2n-  /»a(l  — cosfl) 


/^2  77      />! 

Jo     «/ 0 


n  (1  —  cos  d) 

pdpdO 


for  8(p,  O),  being  constant,  may  be  taken  from  under  the  integral  sign  and 
divided  out.    These  integrals  give  upon  evaluation 

_  _  —  5  aV      3  aV  _      5 
^  ~        4       ""     2     ~  ~  6  ^* 
By  symmetry,  y  =  ^• 

Notice  that  in  all  cases  where  the  density  is  constant  it  divides  out  from 
the  numerators  and  denominators  of  the  expressions  for  the  coordinates  of 
the  center  of  mass. 

(li)  Simple  composite  bodies  composed  of  parts  whose  masses 
and  centers  of  mass  are  already  known.  "We  imagine  the  parts 
replaced  by  particles  of  the  same  masses  situated  at  the  respec- 
tive centers  of  masses,  and  apply  the  formulas  of  §  12.  The  idea 
of  negative  masses  is  helpful  here,  a  negative  mass  corresponding 
to  a  part  C7it  out  from  a  body.  Consider,  for  instance,  a  body  of 
mass  Wj  with  center  of  mass  at  {x^,  y^,  z^  from  which  a  part  m^ 
with  center  of  mass  at  {x^,  y^,z^  is  cut  out;  the  resulting  body 
will  have  a  mass  M—m^—  m,  =  m^  +  (—  m^ ;  let  the  center  of 
mass  be  at  {x,  y,  z). 


36  STATICS 

.   As  the  body  of  mass  m^  may  be  considered  as  composed  of  the 

other  two,  we  have,  by  §  12, 

Mx  +  m„a?„ 

x^  = ?-^> 

M  +  m^ 

in  which,  however,  x  is  the  unknown.    Solving,  we  have 

_  _  (M  +  m^)  x^  —  m,^x^  _  m^x^  —  ''^^^z  _  ''^i^i  +  (~  '^2^2) . 

that  is,  the  same  formula  as  before,  except  that  m^  is  replaced  by 
its  negative. 

Example.  From  a  homogeneous  sj^here  of  radius  a  is  cut  out  a  sphere 
of  radius  b(^a/2)  with  center  at  the  mid-point  of  a  radius  of  the  larger 
sphere.  Find  the  center  of  mass.  Calling  the  density  1  (cf.  the  last 
example),  we  have  a  body  of  mass  tWj  =  4/3  a^  with  center  of  mass  at  the 
origin,  and  a  body  of  mass  —  ?«o  =  —  4/3^'^  with  center  of  mass  at  (a/2, 
0,  0).    Whence 

4  4a 

-      3 3        2  2ah^         _      „       _      „ 

X  — = )     y  =  0,     z  =  v. 

|7ra3-4  7ri3  a^  -  b^       "^ 

If  the  student  desires  to  avoid  negative  masses,  he  may  use  the  formula 
X  =  (m^x^  +  m.^x^)/M,  where  M  is  the  mass  of  the  whole  body,  and  x  the 
a; -coordinate  of  its  center  of  mass.  The  resulting  equation  should  then  be 
solved  for  x-^.    Note  that  m^  =  M  —  m^. 

VI.  PROBLEMS  ON  CENTERS  OP  MASS  OP  CONTmUOUS  BODIES  (ContiiiueO) 

7.  Find  the  center  of  mass  of  a  straight  wire  of  length  I  whose  density 
varies 

(a)  as  the  distance  from  one  end ; 

(b)  as  the  square  of  the  distance  from  one  end ; 

(c)  as  the  nth  power  of  the  distance  from  one  end ; 

(d)  as  (;2-x2); 

(e)  as  sin(7rx/Z); 

(f)  as  sin(7rx/2;). 

8.  Find  the  center  of  mass  of  a  plate  in  the  shape  of  a  qviarter  ellipse 
bounded  by  semimajor  and  semiminor  axes. 

9.  Find  the  center  of  mass  of  a  plate  bounded  by  the  parabola  y"  =  4:ax 
and  the  chord  x  =  h.  Particular  case  where  the  chord  is  the  latus  rectum 
(A  =  a). 

10.  Find  the  center  of  mass  of  three  faces  of  a  cube  regarded  as  plates, 
(a)  when  the  faces  all  meet  in  a  point ;  (b)  when  they  do  not. 


PEOBLEMS  ON  CENTERS  OF  MASS  37 

11.  Find  the  center  of  mass  of  the  cone  formed  by  revolving  y  —  mx 
about  the  x-axis,  between  the  vertex  and  the  base  plane  x  =  h. 

12.  Find  the  center  of  mass  of  the  body  between  the  paraboloid  formed 
by  revolving  y'^  =  ^ax  about  its  axis  and  the  plane  x  =  h. 

13.  Find  the  same  when  the  density  varies  with  \/x. 

14.  Find  the  center  of  mass  of  the  homogeneous  hemisphere  by  method 
of  case  (c). 

15.  Find  the  center  of  mass  of  half  of  the  ellipsoid  of  revolution  obtained 
by  revolving  x^/o^  +  y^/h-  =  1  about  the  x-axis,  the  bounding  plane  being 
the  plants;  =  0. 

16.  Find  the  center  of  mass  of  the  plate  bounded  by  the  x-axis  and  an 
arch  of  the  sine  curve  y  =  sin  x. 

17.  Find  the  center  of  mass  of  a  regular  tetrahedron  composed  of  four 
equilateral  triangular  plates. 

Hint.  Use  oblique  axes  and  regard  the  plates  as  concentrated  at  their 
centers  of  mass. 

18.  Find  the  center  of  mass  of  a  solid  homogeneous  tetrahedron. 

Hint.  Take  origin  at  a  vertex  and  a,-axis  through  the  point  of  intersec- 
tion of  the  medians,  i.e.  at  the  centroid  of  the  opposite  face.  Use  method 
of  case  (d). 

19.  Find  the  center  of  mass  of  a  hemispherical  bowl  of  radius  r. 

20.  Find  the  center  of  mass  of  a  conical  surface  formed  by  revolving 
y  =  mx  about  the  x-axis,  between  the  planes  x  =  a  and  x  —  b. 

21.  Find  the  center  of  mass  of  the  wire  bent  into  the  form  of  a  circular 
arc  of  radius  r  and  angle  2  a.    Check  by  putting  a  —  ir. 

Hint.  Polar  coordinates  advised. 

22.  Find  the  center  of  mass  of  a  wire  bent  into  the  form  of  a  cardioid 
p  =  a(l  —  cos  6). 

Hint.  Use  half  angles. 

2.3.  Find  the  center  of  mass  of  a  spring  x  =  a  cost,  y  =  asint,  z  =  amt, 
the  spring  to  be  ended  by  the  planes  z  =  0  and  z  =  h.  Show  that  the  cen- 
troid lies  on  the  axis  only  when  the  spring  makes  an  integral  number  of 
complete  turns. 

24.  Find  the  center  of  mass  of  the  wire  p  =  e«^  from  ^  =  0  to  ^  =  2  tt. 

2.5.  From  a  right  circular  cylinder  of  height  h  and  base  of  radius  ?•  is 
cut  a  cone  of  the  same  base  and  altitude.  Find  the  center  of  mass  of  the 
remaining  body. 

26.  Find  the  center  of  mass  of  a  hemispherical  shell  bounded  by  a  plane 
and  two  concentric  spheres  of  radii  a  and  b  about  one  of  its  points. 

27.  Show  that  the  center  of  mass  of  the  body  between  two  right  cir- 
cular cones  with  same  vertex,  axis,  and  base  plane,  but  different  angles, 
coincides  with  the  center  of  mass  either  cone  would  have  if  solid. 


38 


STATICS 


A 

A 
1 

1 

C 

Y 

b 

< — ^c--> 

A 

1 

1 

Y 

1 

Y 

Fig.  14 


28.  Find  the  center  of  mass  of  a  plate  in  the  form  of  a  segment  of  a 
circle  of  radius  r,  subtending  an  angle  2a  at  the  center. 

29.  By  adding  the  proper 
triangle  to  the  segment,  get 
the  center  of  mass  of  the  sector 
and  thus  check  the  results  of 
problems  6  and  28. 

30.  Find  the  center  of 
mass  of  the  body  formed  by 
revolving  the  oval  of  y^  = 
x(x-  a)(b-  x),  (6<a<b), 
about  the  x-axis. 

31.  From  a  square  plate 
an  equilateral  triangle  with 
one  side  of  the  square  for  its 
base  is  removed.  Find  the 
center  of  mass  of  the  remain- 
ing plate. 

32.  From  an  elliptic  plate 
x^/a-  +  y^/b^  =  1  a  circular 
plate  of  radius  r  with  center 
at  the  mid-point  of  the  semi- 
major  axis  is  removed.  Find 
the  center  of  mass  of  the 
remaining  plate. 

33.  To  the  elliptic  plate 
of  the  above  problem  are 
attached  two  circular  plates 
of  half  the  thickness  of  the 
elliptic  plate  on  each  side,  so 
as  to  have  the  effect  of  dou- 
bling its  thickness  where 
previously  the  material  was 
removed.  Find  the  center  of 
mass. 

34.  Find  the  center  of  m  ass 
of  a  wire  in  the  shape  of  an 
arch  of  the  cycloid  x  =  o  (^  — 
sin^),  y  =  a(l  —  cos^). 

35.  Find  the  center  of  mass  of  a  triangular  plate. 

Hint.  Use  oblique  axes,  one  axis  in  the  base  of  the  triangle  and  the  other 
the  median  upon  it.  If  rectangular  axes  are  used,  take  the  vertices  at 
(a,  0),  (b,  0),  (0,  c). 

36.  Find  the  center  of  mass  of  the  plate  bounded  by  the  cardioid  p  = 
a(l  —  cos  5). 


);. 

<-- 

1 
1 
1 

1 

a 

> 

1 

b 

< \d — > 

f 

Fig.  15 


PROBLEMS  OX  CENTERS  OF  MASS 


39 


14' 


37.  Find  the  center  of  mass  of  the  plate  bounded  by  half  the  lemniscate 

38.  Find  the  center  of  mass 
of  the  plate  bounded  by  one 
loop  oi  p  —  a  sin  2  6. 

39.  Show  that  the  center  of 
mass  of  the  surface  of  a  zone 
of  a  sphere  is  halfway  between 
the  bounding  planes. 

40.  Find  the  center  of  mass 
of  an  arc  of  the  hypocycloid 
xl  +  yl  =  ai  (or  X  =  a  cos^^,  y  = 
a  siu'<)  between  two  successive 
cusps. 

41.  Find  the  center  of  mass 
of  half  a  right  circular  cone, 
the  dividing  plane  passing 
through  the  axis. 

42.  Find  the  center  of  mass 
of  the  plates  indicated  in  Figs. 
14-17. 

43.  Find  the  center  of  mass 
of  a  wire  in  the  form  of  a  cat- 
enary y=  5(6^/"  +  e--'/")  from 
x  =  0  to  X  =  a. 

44.  Find  the  center  of  mass 
of  a  wire  given  by  the  equa- 
tions y  =  f  Vx^/a,  z  =  X-/4  a 
(or  X  =  afi,  y  =  2  at^/^,  z  = 
ai*/'i),  from  the  origin  to  the 
point  (a,  2  a/3,  a/4). 

45.  Find  the  center  of  mass 
of  a  plate  in  the  form  of  a  sec- 
tor of  the  logarithmic  spiral 
p  =  be"^  bounded  by  the  radii 
^  =^  0  and  ^  =  a. 

46.  Find  the  center  of  mass 

of  the  search-light  reflector  obtained  by  revolving  the  parabola  y^  =  iax 
about  the  x-axis,  and  bounded  by  the  plane  x  =  a. 

47.  Find  the  center  of  mass  of  a  spherical  wedge  bounded  by  two  planes 
meeting  at  an  angle  2  a  and  a  sphere  of  radius  r  with  center  on  their  line 
of  intersection. 

Hint.  Split  up  into  plates  normal  to  edge  of  the  wedge  and  use 
problem  6. 


Fig.  17 


40  STATICS 

48.  Prove  the  ^rst  theorem  of  Pappus:  The  area  of  the  surface  gener- 
ated by  revolving  a  plane  curve  y  =f(x)  between  the  ordinates  x  =  a  and 
X  =  b  about  the  ar-axis  is  equal  to  the  product  of  the  length  of  the  arc 
and  the  length  of  the  path  described  by  the  rotation  about  the  axis  of  the 
arc. 

49.  Prove  the  second  theorem  of  Pappus :  The  volume  of  the  body  within 
the  surface  of  the  above  problem  is  the  area  under  the  curve  times  the 
length  of  the  path  described  by  the  rotation  about  the  axis  of  the  center 
of  mass  of  the  area. 

Hint.  Use  double  integrals. 

50.  Find  by  the  theorems  of  Pappus  the  surface  and  volume  of  the  torus, 
formed  by  rotating  about  an  axis  a  circle  of  radius  a  whose  center  is  a 
distance  b  from  the  axis  (ft>a). 

51.  Find  the  center  of  mass  of  the  part  included  in  a  sjihere  of  radius 
r  and  a  right  circular  cone  whose  elements  make  an  angle  a  with  its  axis, 
and  whose  vertex  is  at  the  center  of  the  sphere. 

52.  Find  the  center  of  mass  of  the  body  bounded  by  a  sphere  of  radius  r 
and  a  right  circular  cone  with  vertex  at  the  center  of  the  sphere,  and  whose 
eleinents  make  an  angle  a  with  its  axis.  Verify  by  putting  a  =  Tr/2  and 
comparing  with  the  result  of  problem  14. 

53.  Find  the  center  of  mass  of  a  spherical  cap  bounded  by  a  sphere  of 
radius  r  and  a  plane  distant  r  cos  a  from  the  center.  Verify  as  in  the  pre- 
ceding problem. 

54.  Find  the  center  of  mass  of  the  part  remaining  of  a  hemisphere  of 
radius  r  after  removing  the  part  contained  within  a  circular  cylinder  of 
radius  a,  and  whose  axis  coincides  with  the  axis  of  the  circle  forming  the 
edge  of  the  hemisphere. 

55.  Find  the  center  of  mass  of  the  plate  bounded  by  the  ar-axis  and  an 
arch  of  the  cycloid  x  =  a(6  —  sin  6),  y  =  a(l  —  cos  6). 

56.  Treat  problems  4  and  5  by  the  method  of  case  (d). 

57.  Find  the  center  of  mass  of  a  parabolic  wire  y"^  =  Aax  bounded  by  the 
latus  rectum.  The  answer  also  holds  for  the  right  cylindric  surface  of 
which  the  wire  is  generatrix,  and  w  hich  is  bounded  by  tw'o  planes  pei"pen- 
dicular  to  the  elements. 

58.  Find  the  center  of  mass  of  the  part  remaining  of  a  sphere  after 
removing  the  part  contained  within  a  cone  with  axis  a  diameter  and  ver- 
tex on  the  surface,  the  elements  of  the  cone  making  an  angle  a  with  its 
axis. 

59.  From  the  problems  on  areas  and  volumes  of  surfaces  of  revolution 
in  your  Calculus  text,  find  by  means  of  the  theorems  of  Pappus  the  center 
of  mass  of  the  corresponding  curves  and  areas  (wires  and  plates). 


ANALYSIS  41 

ANALYSIS  OF  CHAPTER  n 

1.  Subject-matter  of  statics. 

2.  Equilibrium  defined. 

3.  The  two  conditions  for  equilibrium  of  concurrent  forces. 

4.  Analytic  statement  of  the  condition  for  equilibrium  of  concurrent 
forces. 

5.  Assumptions  regarding  frictional  forces. 

6.  Definition  of  the  moment  of  a  force  about  an  axis  ;  its  value  for 
diiferent  positions  of  the  axis. 

7.  Definition  of  couple.    Constancy  of  the  moment  of  a  couple  when 
the  axis  is  shifted  parallel  to  itself. 

8.  The  shifting  of  the  point  of  application  of  a  force  by  the  intro- 
duction of  a  couple. 

9.  Conditions  for  the  equilibrium  of  nonconcurrent  forces. 

10.  Analytic  statement  of  the  first  conditions  for  equilibrium. 

11.  Analytic  statement  of  the  second  conditions  for  equilibrium. 

12.  Definition  of  center  of  mass  and  its  property  with  respect  to  a  system 
of  parallel  forces  proportional  to  the  masses  of  the  parts. 

1-3.  Formulas  for  the  centers  of  mass  of  a  system  of  particles  and  of 
continuous  bodies. 


CHAPTER  III 

DYNAMICS  OF  A  PARTICLE 

15.  Dynamics.  Dynamics  treats  of  the  effect  of  forces  in  produc- 
ing motion.  We  shall  here  limit  ourselves  principally  to  the  motion 
of  particles  in  a  plane,  for  we  thereby  obtain  methods  applicable 
to  the  majority  of  interesting  problems  and  avoid  unnecessary  com- 
plications of  treatment.  Furthermore,  we  start  with  the  simplest 
kind  of  motion  in  a  plane,  namely,  motion  in  a  straight  line. 

16.  Rectilinear  motion;  concepts  involved.  Take  as  origin  a 
convenient  point  0  on  the  line,  and  call  the  distance  OP  of  the 
particle  at  P  from  0,  s ;  then  the  fundamental  notions  involved  in 
the  motion  are  the  distance  s,  the  time  t,  the  velocity  v,  and  the 
acceleration  a*  To  these  should  be  added  the  mass  vi  of  the  body 
and  the  force  /  acting  upon  it.  As,  however,  the  mass  is  constant 
and  /  =  ma,  we  have  really  to  deal  essentially  only  with  the  four 
quantities  above  mentioned,  and  the  problems  of  straight-line 
motion  are  usually  concerned  with  a  relation  between  two  or  more 
of  these.  We  proceed  to  consider  the  more  important  cases  that 
may  arise,  and  to  indicate  how  they  should  be  treated. 

(a)  A  relation  given  between  distance  and  tiine,  say  s  =f{t).  This 
we  shall  call  a  complete  description  of  the  motion,  because  the 

*  Velocity.  If  the  motion  is  uniform,  a  distance  traversed,  .<t2  —  si,  is  proportional  to 
the  time  consumed,  t^  —  ti.  The  velocity  is  then  defined  as  the  constant  ratio  i>  = 
(.92  —  si)/(<2  —  'i)  =  As /At-  If  the  motion  is  not  uniform,  this  ratio  depends  upon  both 
<i  and  to,  and  is  called  the  average  velocity  during  the  interval  t\  to  t^-  By  the  velocity 
at  the  time  t  we  mean  the  limit  of  the  average  velocity  over  an  interval  starting  with 
t  as  the  interval  is  shortened  indefinitely,  i.e.  v  —  Vim  As/ At=  ds/dt. 

Acceleration.  Force  tends  to  change  "motion,"  i.e.  to  change  velocity.  The 
rate  of  change  is  called  acceleration.  If  the  motion  is  "uniformly  accelerated," 
{v2  —  Vi)/{t2  —  ti)=Av/At  is  constant  and  defines  the  acceleration.  If  this  ratio  is 
not  constant,  it  is  called  the  average  acceleration  during  the  interval  t\  to  to.  By 
the  acceleration  at  the  time  t  we  mean  the  limit  of  the  average  acceleration  over  an 
interval  starting  with  t  as  the  interval  is  shortened  indefinitely,  i.e.  a  =  lim  {Av/At) 
—  dv/dt=  ds/dt. 

42 


RECTILINEAR  MOTION  43 

position  of  the  body  is  known  at  every  time,  and  the  questions 
arising,  including  those  concerning  velocity  and  acceleration,  are 
answered  by  means  of  differentiations.  All  other  cases  involve  inte- 
grations. Moreover,  in  this  case  only  does  the  given  relation  suf- 
fice to  determine  uniquely  the  motion,  for  there  are  no  constants  of 
integration  here.  The  prohlem  in  any  other  case  may  he  considered 
as  essentially  solved  by  reduction  to  this  one;  that  is,  hy  deriving  a 
relation  ^  _  ^/-^\ 

Illustrative  Exercise.  Consider  the  motion  s  =  t^—'&t.  Find  v 
and  a.  When  does  the  body  come  to  rest,  and  at  what  points?  How 
does  it  move  before,  between,  and  after  these  times  of  rest?  When  is 
the  force  to  the  right  and  when  to  the  left?    Discuss  similarly  the  motion 

(b)  A  relation  given  between  acceleration  (or  force)  and  time,  say, 
a  =f{t).  To  reduce  this  back  to  the  complete  description  of  case  (a) 
requires  two  integrations.    As  a  =  dv/dt,  we  have 

dt 
and  as  ■?;  =  dsfdt,  we  have 

—  =  I  f{t)  dt  +  Cj,     whence      ^—\\  f(^)  ^^^^  +  ^i^  +  ^2- 

The  constants  cannot  be  determined  without  auxiliary  conditions,' 

which  usually  accompany  eacli  problem.     They  will  always  be 

given  in  the  exercises  below. 

Example.  A  body  is  repelled  from  a  fixed  point  by  a  force  proportional 
to  the  time  t.  The  body  starts  from  rest  at  the  origin  when  /  =  0.  Find 
the  complete  description.  The  conditions  of  the  problem  are/=  ^7,  whence 
ma  =  kt;  and  ;;  =  0,  .s- =  0,  when  <  =  0.  The  first  gives  dv/dt  =  {k/vi)t, 
whence  v  =  {k/2  m)  l^  +  Cj.  But  as  y  =  0  when  t  =  0,  we  have  0  =  0  +  Cj. 
Whence  Cj  =  0  and  y  =  (k/2  m)  t^.  But  v  =  ds/dt.  Whence  ds/dt  =  (^•/2  m)  t^ 
and  s  =  {k/Q  m)t^  +  c^.  But  as  *•  -  0  when  t  =  0,  we  have  c^=  0,  whence 
s  —  (I'/G  m)t^,  the  required  relation. 

(c)  A  relation  given  between  acceleration  {or  force)  and  distance, 
say,  a=f(s).  This  also  requires  two  integrations,  which  may  be 
carried  out  as  follows :  Write  the  relation  dv/dt  =/(«),  and  multiply 


=  f{t),     whence     v  =  I  f{t)dt  +  e^, 


44  DYNAMICS  OF  A  PARTICLE 

by  2  -y  on  the  left  and  by  its  equivalent  2  (ds/dt)  on  the  right, 
obtaining  2v(dv/dt)=  2/{s){ds/dt).  Whence,  integrating  with 
regard  to  t,  we  have 

v'=  f2f{s)ds  +  c^. 

To  carry  out  the  second  integration,  solve  for  v,  obtaining 


-^-H' 


f(s)ds  +  c^, 


whence    dt  =  — p==^=;  and  t  =  j  — t^====  +  c^. 
y^2Jf{s)  ds  +  c,  j  ^2Jf{s)  ds  +  c. 

If,  after  determining  c^  and  c^  by  the  auxiliary  conditions  given, 
and  carrying  out  the  integrations,  the  resulting  equation  is  easily 
solvable  for  s,  we  have  our  complete  description  of  case  (a).  Other- 
wise we  have  a,  v,  and  t  all  expressed  as  functions  of  s,  which  may 
be  considered  as  giving  a  parameter  representation  of  the  motion. 

Example.  A  body  falls  toward  the  origin  under  a  force  inversely  pro- 
portional to  the  square  of  the  distance  away.  If  v  ajsproaches  0  as  s 
approaches  co,  and  if  .<?  =  0  for  t  =  0,  determine  the  motion.  Considering 
the  body  to  move  on  the  positive  part  of  the  line,  the  force  will  be  nega- 
tive, as  it  is  toward  the  origin.  A  proportionality  factor  giving  simple 
results  is  —  km/2.  Then,  as/=  —  hn/2s^,  and  ma  =/,  we  have  a  =  dv/dt 
-  -  Ic^/I  s2.  Then,  as  above,  2  v  {dc/dt)  =  -  (F/*-2)  {ds/dt) ,  or  v^=k'^/s  +  c^. 
But  w  =  0  as  s  =  00,  hence  0  =  0  +  Cj  and  r^  =  k'^/s,  whence  v  —  ±  k/y/s. 
The  sign  to  choose  is  the  negative  one,  as  the  body  is  falling  toward  the 
origin.  Hence  v  =  ds/dt  =  —  k/Vs  and  Vsds  =  —  kdt.  Hence  |  s'=  —kt+c„, 
where  0^=0  by  the  auxiliary  conditions.    Thus  the  motion  is  given  by 

3  kty 

Among  some  interesting  questions  that  may  be  attached  to  the  above 
problem  are  :  (1)  Has  the  body  fallen  from  an  infinite  distance  in  finite 
time?  (2)  With  what  velocity  does  it  reach  the  origin?  (3)  Does  the 
solution  given  hold  for  positive  or  negative  t?  (4)  Has  the  solution  a 
meaning  when  t  has  the  other  sign,  and  if  so,  what  is  it? 

(d)  A  relation  given  between  acceleration  (or  force)  and  velocity, 
say,  a  =f(y).    Two  integrations  are  necessary.    First 


RECTILINEAR  MOTION  45 

-r  =/('^)>     whence     /— —  =  ^  +  c,. 

This  relation  must  then  be  solved  for  v,  giving,  say,  v  =  F(t  +  c^). 
Thence  ^ 

s=      F(t  +  c^)dt+c^. 

A  relation  between  the  velocity  and  the  distance  also  may  be 
found  by  one  integration.    Thus,  as 

_  dv      dv    ds  _     dv 
dt       ds    dt         ds 

we  have  v---  =  f(v),     whence     i  — —  =  s  +  c. 

ds      '^^^'  J  f(v) 

The  formula  a  =v  (dv/ds)  is  a  very  useful  one. 

(e)  A  relation  given  connecting  acceleration  (or  force),  velocity, 
and  distance.  In  the  more  interesting  problems  this  relation  is 
linear  in  a,  v,  and  s,  and  has  constant  coefficients,  say,  a  -f-  hv'+  cs  =  0 
or  d%/dt^  +  b  (ds/dt)  +  cs  =  0,  for  the  integration  of  which  the 
student  is  referred  to  text-books  on  the  calculus  or  on  differential 
equations. 

Graphic  representations  of  a  motion,  (a)  On  a  straight  line.  Plot  the  points 
corresponding  to  a  succession  of  values  of  t  differing  by  equal  increments. 
The  direction  of  motion  may  be  indicated  by  arrows  parallel  to  the  line. 
The  closeness  together  of  the  points  gives  an  indication  of  the  speed.  If  the 
motion  returns  upon  itself,  draw  a  number  of  parallel  lines  and  on  each 
one  plot  the  motion  between  two  of  its  turning  points.  If  in  doing  this  we 
imagine  the  points  actually  marked  at  equal  intervals  of  time,  the  point  of 
the  pencil  has  a  motion  which  is  a  picture  of  the  required  motion. 

(b)  Using  the  plane.  Draw  a  <-axis  perpendicular  to  the  s-axis  and 
plot  s  =f(t).  If  now  a  slit  be  cut  in  a  piece  of  paper,  and  the  plot  be  put 
behind  the  slit,  so  that  the  slit  is  parallel  with  the  s-axis,  and  if  the  plot 
be  drawn  with  uniform  speed  downward  in  the  direction  of  the  negative 
Z-axis,  the  point  of  the  plot  which  shows  through  the  slit  will  appear  with 
the  required  motion.  The  maxima  and  minima  of  s  =f(t)  will  be  the  turn- 
ing points,  and  the  speed  is  given  by  the  slope  of  the  curve  with  respect 
to  the  /-axis.  In  discussing  the  motion  the  student  should  use  one  of  these 
methods,  and  it  is  highly  valuable  for  him  to  become  acquainted  with  both 
(see  Fig.  18). 


46  DYNAMICS  OF  A  PARTICLE 

VII.   PROBLEMS  ON  RECTILINEAR  MOTION 

In  the  first  few  of  the  following  problems  the  "  complete  description  "  is 
given  and  the  motion  should  be  discussed.  A  discussion  should  include 
such  points  as  the  following  : 

(a)  Times  and  points  where  the  motion  stops,  i.e.  where  the  velocity 
vanishes. 

(b)  The  points  reached  by  the  motion  ;  e.g.  in  s  =  fi  only  the  points  to 
the  right  of  the  origin,  with  the  origin. 

(c)  Direction  of  the  motion  between  stops  (judged  by  the  sign  of  v). 

(d)  The  number  of  times  each  part  of  the  line  is  traversed,  e.g.  in  the 
above  example  the  positive  half  of  the  line  is  traversed  twice,  the  negative 
half  not  at  all. 

(e)  Direction  of  the  acceleration  (i.e.  of  the  force). 

(f)  Tendency  of  the  motion  for  large  negative  and  for  large  positive 
values  of  t. 

It  may  be  that  not  all  these  characteristics  will  be  of  interest  in  a  given 
motion,  whereas  special  motions  will  have  other  characteristics  of  special 
interest,  —  which  should  be  pointed  out.  The  student  should  in  advance 
get  clearly  in  mind  the  meanings  of  sign  of  the  velocity  and  acceleration. 
This  may  be  done  by  answering  the  question.  What  is  characteristic  in  the 
following  four  types  of  motion  :  (l)y;>0,  a>0;  (2)  v>0,  a<0  ;  (3)  i;<0, 
rt>0;   (4)  i;<0,  a<0? 

1.  s  =  fi-Qfi  +  10. 

Solution.  v  =  St^-12i  =  ^t(t-^).  a  =  6  t  -  12  =  6(t  -  2).  (a)  The 
motion  ceases  for  t  =  0  and  t  =  4,  that  is,  at  the  points  s  =  10  and  s  =  —  22. 

(b)  All  points  are  reached,  because  a  cubic  equation  t^  —  G  t^  +  10  =  s 
will  have  a  real  root,  no  matter  what  the  value  of  s. 

(c)  The  direction  of  motion  depends  iipon  the  sign  of  v,  which  depends 
upon  the  signs  of  its  factors.  For  t<0,  v>0,  and  the  motion  is  forward  to 
s  =  10.  For  0<t<4,  v<0,  and  the  motion  is  backward  to  —  22.  For  4<<, 
v>0,  and  the  motion  is  forward  always  thereafter. 

(d)  The  points  between  —  22  and  +  10  are  traversed  three  times,  other 
points  once. 

(e)  The  acceleration  is  backward  till  t  =  2,  at  the  point  s  =  —  6,  then 
forward. 

(f)  The  farther  back  in  time  we  go,  the  farther  to  the  left  was  the  point 
(because  for  large  negative  t,  s  is  large  and  negative),  and  there  is  no  point 
to  the  left  of  which  the  moving  point  has  not  been.  The  velocity  is  positive 
for  negative  t,  but  decreasing  as  time  progresses  ;  that  is,  the  motion  is  a 
slower  and  slower  forward  motion.  For  larger  and  larger  positive  t  we  find 
s  increasing  without  limit  and  v  also  ;  that  is,  the  point  moves  forward 
beyond  any  point  whatever  with  an  always  increasing  velocity. 

Graphic  representations  of  the  motion  are  shown  in  Fig.  18. 


PROBLEMS  ON  RECTILINEAR  MOTION 


47 


2.  A-  =  -  16  e  -  32 1 

-10. 

12. 

s  =  log  t. 

3.  s  =  ht  +  SQ. 

13. 

s  =  e-*  •  sint. 

4.  s  =  2<8- 3/2-1 

2<  +  12. 

14. 

s  =  e--^/'\ 

5.  s  =  l/L 

15. 

s  =  h(t-iy. 

6.  s  =  l/(<2  +  l). 

16. 

s  =  b  ■  tan  t. 

7.  s  =  l/02_i). 

17. 

s  =  lie^'  +  e-kty 

8.  s  =  6sin^ 

18. 

s  =  l(e'--t-e- ''■'). 

9.  s  =  6  sin  (A-/ +  e). 

19. 

s  =  fc(ci'  -  e-^')/(e*«  +  e-*'). 

10.  s  =  e'. 

20. 

/  =  s»-6s-l. 

11.  s  =  e-'. 

21. 

/  =  s  +  sins. 

Can  a  point  actually 

have  the 

motion 

given  by  problems  20  and  21  ? 

Fig.  18 


In  the  next  problems  find  the  "  complete  description,"  determining  the 
constants  of  integration  by  the  auxiliary  conditions  given.  If  the  motion 
is  not  recognized  as  one  of  the  above  types,  it  should  also  be  discussed  in 
the  same  way.    Tell  in  all  cases  the  force  involved. 


48  DYNAMICS  OF  A  PARTICLE 

22.  I'  -  kt;  s  =  l  Avhen  t  =  0. 

23.  V  =  ks ;  s  =  Sf^  when  t  =  t^^. 

24.  a  =  kv;  v  =  2,  s  =  l  when  t  =  0. 

25.  a  =  A:< ;  s  =  0  when  t  =  0,  s  =  0  when  <  =  1. 

26.  a  =  ^-^s ;   y  =  0,  s  =  o  when  t  =  0. 

27.  a  =  /j% ;  v  =  ka,  s  =  0  when  t  —  0. 

28.  a  =  —  k^s;  v  =  v^,  s  =  0  when  t  =  <„. 

29.  a  —  —  k^/s^;  v  approaches  0  as  s  approaches  co,  s  =  s^  when  t  =  0. 
Show  that  in  this  problem  the  body  can  only  move  on  the  positive  side  of 
the  origin. 

30.  a  =  —  2  bv  +  cs ;  obtain  the  general  solution  and  discuss  for  b'^  ~  c% 0. 

17.  Some  special  rectilinear  motions,  (a)  The  inclined  plane. 
Let  i  denote  the  angle  of  inclination  of  the  plane  to  the  horizon. 
Then,  resolving  the  force  of  gravity  into  two  components,  perpen- 
dicular and  parallel  to  the  plane  respectively,  the  parallel  compo- 
nent, mg  cos  i,  is  alone  effective  in  producing  motion.  The  further 
development  of  this  topic  is  left  to  the  student  in  the  exercises 
and  problems  to  follow. 

Ex.  1.  Determine  the  motion  of  a  body  sliding  from  rest  down  an 
inclined  plane  of  inclination  i. 

Ex.  2.  Show  that  bodies  starting  simultaneously  to  slide  down  various 
chords  of  a  vertical  circle  from  the  highest  ix)int  of  the  circle  will  all  reach 
the  circumference  at  the  same  time. 

(b)  Simple  harmonic  motion.  Let  a  body  be  attracted  toward  a 
fixed  center  with  a  force  varying  with  its  distance  away  from  the 
center,*  Let  the  proportionality  factor  be  mk^ ;  then,  as  the  sign 
of  the  acceleration  is  opposite  to  that  of  s,  we  have  a  =—  l^s,  or 
d^s/df+  ^s  =  0.  If  this  be  integrated  as  a  linear  equation  with 
constant  coefficients  (see  also  the  exercise  below),  we  have  for  the 
solution  s  =  c^  cos  kt  +  c^  sin  kt.  It  is  shown  in  Analytic  Geometry 
that  two  numbers  c^  and  c^  are  always  proportional  to  the  sine 
and  cosine  of  some  angle  e,  say,  c^=  A  sin  e,  c^=  A  cos  e.  Then  s 
becomes  s  =  ^  sin  {kt  +  e).    The  student  should  verify  by  direct 

*  Examples  of  forces  of  this  sort  are  elastic  forces,  for  small  displacements,  due 
to  springs  and  elastic  bands;  gravity  in  mines  and  beneath  the  surface  of  the  earth; 
and,  approximately,  the  forces  involved  in  pendulum  motion  and  the  motion  of  mag- 
netic needles. 


K^,+  e  =  K+e  +  2'rr,     or     t^^^-t^^T  = 


SPECIAL  EECTILIXEAR  MOTIONS  49 

substitution  that  this  satisfies  the  differential  equation.  This 
motion  the  student  has  studied  in  problem  9  of  the  previous  par- 
agraph.   The  motion  repeats  itself  at  intervals  such  that 

2j7r 
k  ' 

This  quantity  T  is  called  the  period  of  the  simple  harmonic 
motion. 

Ex.  3.  Integrate  the  above  differential  equation  by  the  method  of  case 
(c)  in  §  16  and  reconcile  the  results. 

Ex.  4.  Discuss  dampened  harmonic  motion  in  which  there  is  a  resistance 
due  to  air  or  friction  proportional  to  the  velocity,  so  that  a  =  —  kh  —  bv. 
Compare  with  problem  13  of  §  16,  which  is  a  special  case. 

(c)  Fall  of  a  body  from  a  great  height.  The  force  of  gravity 
acting  upon  a  body  is  sensibly  constant  if  the  distance  through 
which  the  body  moves  is  small.  Careful  measurements,  however, 
do  reveal  a  variation,  and,  according  to  Newton's  law  of  universal 
gravitation,  the  force  upon  a  body  outside  the  earth's  surface  varies 
with  the  inverse  square  of  the  distance  from  the  earth's  center. 

Let  us  take  for  origin  the  point  from  which  the  body  falls,  say, 
at  a  height  h  from  the  earth's  center,  and  let  the  direction  toward 
the  center  be  taken  as  positive.  Then  h  —  s  is  the  distance  of  the 
body  from  the  center,  and  /  =  ma  =  mc/{h  —  s)^.  If  R  denote  the 
radius  of  the  earth,  we  have,  since  at  the  surface  a  =  g,  g  =  c/R^, 
or  c  =  gR^.    Thus  our  equation  becomes 

_g^_  dh  __     gR^ 

""-{h-sf     ""'     df~{h~sf 

which  comes  under  case  (c)  of  the  last  paragraph.  We  have,  then, 

dv        ,-.2       1       ds  ,  v^       ,     qR^    , 

v-r  =  qR  — 1  —  >     whence     —  =  +  -r- h  c,. 

dt      ^      {h-sfdt  2  h-s       ' 

As  the  body  falls  from  rest,  v  =  0  when  s  =  0,  so  that 

and  therefore  v''  =  2  gR^  (-^ l\=2  gR'  (ttt—-)  • 

^      \h  —  s      hj         -^      \h{h  —  s)l 


50  DYNAI^IICS  OF  A  PARTICLE 


Whence  '^  =  -7-  =  -^  v  ,       v  7 

(it  ^  11      ^h 


^¥-^ 


This  admits  of  further  integration  by  the  ordinary  methods,  but 
the  integral  is  more  useful  when  obtained  in  the  form  of  a  series. 
We  have  , — 

For  the  motion  which  interests  us  0  <  s  <  h,  so  that  we  develop 
the  radical  as  a  series  in  s/h. 


1/A     ^-l/sV    \-l-^/s 


1-^  ;  - 


h      _ 

s)  [^      2\hJ      l-2\h/      1.2- 3  VA 
hV     l/s\^     1    l/s\i     1    1    3/s 


sj      2\hJ       2    4:\h/       2    4    6\^ 

Using  this  series  in  the  differential  equation  and  integratmg,  we 

have  

1     |/'   /0/7  x>      2s?      1    1    2.si      1    1    3   2s?  \ 

BV2ff     [        h      2    4    5\hJ      2    4    6    7^^ 

where  we  have  put  c^=0  on  the  supposition  that  s  =  0  when  ^  =  0 ; 
tliat  is,  that  we  count  the  time  from  the  instant  the  body  starts  to 
fall.  If  s  is  small  compared  with  /*,  this  series  converges  and  a 
few  terms  give  a  good  approximation  to  the  true  motion. 

Ex.  5.  Write  out  the  nth  term  of  the  above  series. 

Ex.  6.  Taking  simply  the  first  term  of  the  series  above  and  putting 
h  =  R,  show  that  we  have  the  ordinary  law  for  falling  bodies  at  the  earth's 
surface. 

.Ex.  7.  The  terms  of  the  series  after  the  first  are  all  negative,  so  that 
the  actual  motion  differs  from  that  considered  in  the  preceding  exercise  in 
a  way  easy  to  state.    Make  the  statement. 

(d)  Bodi/  falling  through  a  resisting  medium.  Raindrops  and  me- 
teorites are  examples  of  this  case ;  similar  forces  act  upon  sliips 


SPECIAL  KECTILINEAR  MOTIONS  51 

and  trains  moving  with  small  velocities  under  propulsion  of  their 
own  engines.  The  resistance  we  shall  assume  to  be  proportional 
to  the  square  of  the  velocity,  and  take  our  s-axis  downward.  We 
have,  then,  n 

where  g  is  the  acceleration  due  to  gravity  and  g/Jc^  a  con^^enient  pro- 
portionality factor.    The  differential  equation  is,  writing  a  =  dv/dt, 


dv  g 


.^dt, 


whence     — -  log =  —  ^t-{-  const.,     or     =  c.e    *  . 

This  shows  clearly  a  tendency  of  the  motion,  namely,  as  time  goes 
on  e-<:2!//'f)«  grows  rapidly  small,  and  hence  v  approaches  k,  that  is, 
the  velocity  tends  to  become  constant.  The  constant  k,  introduced 
above,  thus  receives  as  its  interpretation  this  limiting  velocity.  If 
the  body  fall  from  rest,  we  have  v  =  0  for  ^  =  0,  so  that  —  1  =  c^, 
and  using  this  value  and  solving  for  v,  we  have 

2  fjt  gt  gt 

ds      ,  g  *  —  1       ,  e^  — g   * 
V  =  —  =  A:  • =  k > 

whence  s  —  ■?„  =  —  log 


9 

As  noted  before,  er'^'J'^'^^  tends  toward  zero  as  t  increases,  and  we 
obtain  an  approximation  by  dropping  it.  The  result  of  this  approx- 
imation is  the  uniform  motion  s  =  kt  -\-  const. 

Ex.  8.  Show  that  if  the  body  have  a  downward  initial  velocity  of  v^,  the 
integral  will  be  s  —  .s^  =  ^  log [i'q  sinh  {g/k) t  +  k  cosh {g/k) t\  where 

gi< g—  u  e"  +  e"  " 

sinh  u  — ,     and     cosh  u  = • 

2  2 

Ex.  9.  Study  the  motion  of  a  body  on  which  no  constant  force,  but  the 
resistance  alone,  acts,  there  being  an  initial  velocity  v^. 

*  This  equation  holds  only  during  downward  motion,  for  if  the  velocity  were 
upward,  i.e.  negative,  the  resistance  is  added  to  the  acceleration  due  to  gravity.  For 
upward  motion  the  law  \sa  =  g  +  gv^/k^.  The  student  should  hear  in  mind  that  any 
simple  law  of  resistance  like  the  ahove  is  merely  an  approximation.  See  Osgood, 
Differential  and  Integral  Calculus,  p.  21G. 


62  DYNAMICS  OF  A  PARTICLE 

Vin.  PROBLEMS  ON  BODIES  MOVING  IN  STRAIGHT  LINES  UNDER  THE 
ACTION  OF  GIVEN  FORCES* 

1.  A  body  falls  from  a  height  100  m.  After  falling  for  2  sec.  a  second 
body  is  projected  vertically  upward  from  the  earth  toward  the  first  with  a 
velocity  40  m.  per  sec.    Find  the  time  and  height  at  which  they  meet. 

Solution.  Let  us  count  the  time  from  the  instant  the  first  body  begins 
to  fall,  and  the  distance  vertically  upward  from  the  earth.  If  we  use  sub- 
scripts to  distinguish  the  two  bodies,  we  then  have  the  conditions  a.^=  —  g, 
with  Tj  =  0,  Sj  =  10,000  when  t  =  0  for  the  first  body,  the  units  being  centi- 
meters and  seconds  ;  and  a^=  —  g,  with  v^  —  4000  and  s^  —  0  when  <  =  2. 
These  give  s^  =  -  gfi/2  +  10,000,  s^  =  -  g\t  -  2)72  +  4000  {t  -  2).  The 
bodies  meet  w  hen  s^  —  s^  ;  equating  the  two  ex])ressions,  we  find  the  time  of 
meeting  to  be  <  =  3.35  sec.  nearly,  and  this  gives  s^  =  Sg  =  4500  cm.  or 
45  m.  about. 

Remarks,  (a)  It  is  of  highest  importance  that  we  be  consistent  in  our  use 
of  units.  In  the  C.G.S".  system  everything  shoidd  be  reduced  to  centimeters, 
grams,  and  seconds.  The  value  of  g  is  approximately  981  cm.  jjer  sec. 
per  sec. 

(b)  Our  problem  is  usually  in  these  exercises  to  determine  the  force  act- 
ing on  a  body,  to  equate  it  to  the  mass  times  the  acceleration,  and  to  deter- 
mine the  initial  conditions.  Thence  we  determine  the  complete  description 
of  the  motion  and  by  means  of  it  answer  the  questions  proposed. 

2.  A  balloon  is  ascending  with  a  velocity  20  mi.  per  hr.  when  a  stone  is 
dropped  from  it.  The  stone  reaches  the  ground  in  6  sec.  Find  the  height 
of  the  balloon  when  the  stone  was  dropped. 

3.  How  high  will  a  stone  rise  if  thrown  upward  with  a  velocity  80  m. 
per  sec.  ? 

4.  Show  that  a  body  thrown  upward  has  the  same  speed  in  ascending 
and  descending  past  a  given  point. 

5.  A  body  thrown  into  the  air  with  a  velocity  jJq  attains  a  certain  height 
before  falling.  How  much  must  the  velocity  be  increased  in  order  to  double 
this  maximum  height? 

6.  A  weight  12  lb.  on  an  inclined  plane  of  inclination  i  =  arctan  1/2  is 
connected  by  a  string  to  a  weight  8  lb.  hanging  over  the  upper  edge  of  the 
plane  and  starts  from  rest.    Find  the  distance  described  in  5  sec. 

7.  A  body  weighing  9  kgm.,  on  a  smooth  table,  3  m.  from  its  edge  is 
connected  by  a  string  to  a  body  weighing  1  kgm.  and  hanging  over  the  edge. 
Find  (a)  their  common  acceleration,  (b)  the  time  when  the  first  mass 
leaves  the  table,  (c)  its  velocity  upon  leaving. 

8.  Two  particles  of  mass  m^  and  7n^  are  connected  by  an  inextensible 
string  which  passes  over  a  frictionless  pulley.    If  m^>m^,  show  that  the 

*  Except  when  otherwise  stated,  the  force  of  gravity  is  to  be  considered  constant 
and  resistances  are  to  be  neglected. 


PHOBLEMS  ON  EECTILINEAR  MOTION  53 

tension  T  on  the  string  is  2  m.^m^/(m^  +  m^)  and  determine  the  motion  of 
the  system.  (Note  that  if  Sj  and  s^  are  the  distances  of  the  particles  below 
the  pulley,  s.^  +  s^  =  constant  and  hence  v^  +  Wg  =  «!  +  ^2  ~  ^O 

9.  Show  that  diiring  the  motion  in  the  above  problem  the  pressure  on 
the  axle  is  less  than  the  sum  of  the  weights  of  the  two  bodies.  (Be  careful 
to  express  all  forces  in  the  same  units.  The  theorem  may  be  proven  by  the 
fact  that  a  perfect  square,  (wjg  —  niiY,  is  positive.) 

10.  Over  a  pulley  passes  a  string  to  one  end  of  which  is  attached  a 
weight  10  lb.,  and  to  the  other  a  weight  8  lb.  with  a  rider  4  lb.  After 
being  in  motion  5  sec.  the  rider  is  removed  without  checking  the  velocity. 
How  much  farther  will  the  system  move? 

11.  With  what  velocity  must  a  particle  be  projected  downward  in  order 
to  overtake  in  10  sec.  a  body  that  has  already  fallen  100  ft.  from  rest  at 
the  same  point  ? 

12.  Given  a  point  and  a  vertical  line  a  distance  d  from  it,  find  the 
inclination  of  the  straight  line  which  would  guide  a  particle  acted  upon  by 
gravity  only,  from  the  point  to  the  line  in  the  briefest  time. 

13.  Consider  the  same  problem  when  the  line  is  not  vertical. 

14.  AVhat  is  the  weight  by  a  spring  balance  of  a  man  of  160  lb.  descend- 
ing in  an  elevator  with  an  acceleration  2  ft.  per  sec.  per  sec.  ? 

15.  A  high  JTimper  in  jumping  raises  his  center  of  mass  3  ft.  How  high 
could  he  jump  on  the  surface  of  the  moon,  where  he  weighs  one  sixth  as 
much?  How  long  is  he  off  the  ground  in  both  cases?  (Assume  he  leaves 
the  ground  with  the  same  velocity  in  both  cases.) 

16.  If  a  motor  car  with  speed  40  mi.  per  hr.  can  be  stojiped  by  its 
brakes  in  a  hundred  yards,  find  the  inclination  of  a  hill  on  which  the 
brakes  would  just  hold  it. 

17.  A  train  with  velocity  30  mi.  per  hr.  runs  with  steam  shut  off 
against  a  resistance  of  10  lb.  i^er  T.    How  far  will  it  go? 

18.  A  train  running  at  15  mi.  per  hr.  with  steam  shut  off  strikes  an 
up  grade  of  1  in  300.  The  resistance  due  to  friction  and  air  averages  3  lb. 
per  T.    Find  how  far  the  train  will  run  before  stopping. 

19.  A  train  of  weight  200  T.  descends  a  grade  of  inclination 
i  =  arcsin^^^.  If  its  velocity  is  initially  40  mi.  per  hr.,  what  frictional 
resistance  will  stop  it  in  half  a  mile? 

20.  The  attraction  of  the  earth  for  a  particle  of  mass  m  beneath  the 
surface  of  the  earth  varies  as  the  distance  from  the  earth's  center,  and  it 
is  mg  at  the  surface.  Find  how  soon  the  body  would  reach  the  center  if 
dropped  into  a  hole  through  the  center  ;  also  its  maximum  speed,  and  at 
what  point  this  speed  is  attained. 

21.  An  elastic  band  is  stretched  between  two  points  on  a  smooth  table. 
A  weight  fixed  at  the  mid-point  of  the  band  moves  in  the  line  of  the  band, 


54  DYNAMICS  OF  A  PARTICLE 

the  elastic  force  being  proportional  to  the  displacement.   Discuss  the  motion 
from  its  equations. 

22.  To  one  end  of  an  elastic  cord  of  natural  length  2  ft.  is  attached  a 
weight  2  lb.,  which  when  hanging  in  equilibrium  extends  the  cord  to  a 
length  2^  ft.  Assuming  Hooke's  law,  that  the  elastic  force  exerted  b}^  the 
cord  is  proportional  to  its  increase  in  length,  determine  how  far  the  weight 
would  fall  if  allowed  to  drop  from  the  point  at  which  the  suspending  cord 
has  its  natural  length  2  ft. 

23.  A  cylinder  of  radius  5  cm.  is  weighted  at  one  end  so  as  to  float 
vertically  in  water.  If  forced  downward  into  the  water  a  distance  5  cm. 
below  its  position  of  equilibrium  and  then  released,  it  is  found  to  oscillate 
vertically  with  a  period  2  sec.    Determine  the  cylinder's  weight. 

24.  Find  the  velocity  with  which  a  body  reaches  the  earth's  surface  in 
falling  from  a  height  equal  to  the  earth's  radius,  resistance  of  the  atmos- 
phere being  neglected.  Find  the  time  occupied  by  the  fall.  (Use  aboiit 
five  terms  of  the  series.) 

25.  As  we  increase  the  distance  through  which  a  body  falls  to  the 
earth,  show  that  the  velocity  with  which  it  reaches  the  surface  approaches 
an  upper  limit  ^2  gR.  If  a  body  were  shot  upward  with  this  velocity,  what 
would  happen? 

26.  A  ship  steaming  at  its  maximum  rate  of  12  mi.  per  hr.  is  stopped 
by  reversed  engines  and  resistance  in  6  sec.  What  is  the  value  of  k,  and 
what  is  its  velocity  after  2  sec.  ?  (See  footnote,  p.  51.) 

27.  In  the  motion  studied  under  (d)  of  the  present  section,  character- 
ized by  the  equation  ma  =  m  (^g  —  g /  k'^v^) ,  the  coefficient  of  resistance 
c  =  mg/k^  depends  only  upon  the  size  and  shape  of  the  body,  and  not  upon 
its  mass.  Using  this  fact,  compare  the  ultimate  speeds  k^  and  k^  acquired 
by  a  falling  raindrop  and  a  falling  bullet  of  the  same  size,  the  specific 
gravity  of  lead  being  11.3. 

28.  In  problem  21  suppose  the  body  experience  a  resistance  equal  to 
2  kv  as  well  as  the  elastic  force.  Show  that  the  motion  becomes  a  dampened 
harmonic  one  of  the  type  s  =  Ae-^' sin  (fit  +  B),  A  and  B  being  constants 
of  integration. 

29.  Study  the  motion  of  a  particle  repelled  from  a  center  by  a  force 
proportional  to  the  distance  of  the  particle  from  the  center.  Such  a  force 
might  arise  in  studying  electrified  bodies. 

30.  A  chain  of  leng-th  5  ft.  rests  upon  a  smooth  table  with  1  ft.  of  its 
length  hanging  over  the  edge.  Determine  the  motion  of  the  chain  as  it 
slides  off ;  also  the  time  when  it  leaves  the  table  and  its  velocity  at  this 
moment. 

Hint.  Treat  the  problem  as  if  the  chain  were  falling  vertically,  but  only 
the  weight  of  the  part  of  the  chain  below  the  level  of  the  table  were  effect- 
ive in  producing  motion. 


PLANE  MOTION 


55 


motion 


18.  Motion  of  a  particle  in  a  plane.  In  taking  up  motion  in  the 
plane  it  is  essential  to  recall  that  velocity,  acceleration,  and  force 
are  iw  longer  mere  numbers,  but  directed  magnitudes,  that  is,  vectors. 
We  shall,  as  a  rule,  think  of  them  as  fixed  by  their  two  projections 
on  the  axes.  We  consider  first  a  complete  description  of  the  motion 
(cf.  p.  42).  This  wiU  evidently  be  attained  if  we  know  the  coor- 
dinates of  the  point  at  any  given  time ;  in  other  words,  the  complete 
description  will  consist  in  two  equations  :* 

x=f{t)\ 

y  =  9it)r 

It  is  thus  seen  that  the  complete  description  of  a  plane  motion 
j  consists  in  the  description 
of  two  straight-Hne  motions, 
namely,  those  of  a  point  on 
the  a,'-axis  directly  beneath  ^ 
(or  above)  the  point  in  the  ^ 
plane,  and  of  a  point  on  the 
y-axis  on  a  level  with  it. 
This  dependence  of  the  mo- 
tion of  a  point  in  the  plane 
upon  two  straight-line  mo- 
tions is  fundamental,  as  our 
whole  analytic  treatment  of 
plane  motion  depends  upon  it,  and  a  similar  statement  holds  for 
space. 

The  coordinates  x  and  y  of  the  moving  point  may  be  considered 
the  projections  of  a  vector  whose  beginning  is  at  the  origin  of 
coordinates  and  whose  end  is  at  the  moving  point;  we  shall  call 
it  the  position  vector.  We  now  define  the  velocity  vector  as  the 
derivative  with  respect  to  the  time  of  the  position  vector,  and  simi- 
larly we  define  the  acceleration  vector  as  the  derivative  with  respect 

*  Another  form  of  description  sometimes  of  value  consists  in  the  equation  of  the 
path  f(x,  y)  =  0,  or  a;  —f{s),  y  =  g (s),  together  with  a  relation  showing  how  the  path 
is  described,  say  s  =  4>{t),  where  s  represents  the  length  of  the  path  measured  from 
some  convenient  point  on  it.  The  question  of  passing  from  one  kind  of  description  to 
the  other  is  left  for  the  student  to  consider. 


x-At) 

Fig.  19 


66 


DYNAMICS  OF  A  PARTICLE 


to  the  time  of  the  velocity  vector;*  or,  referring  to  §§  5  and  6,  the 
velocity  vector  V  and  the  acceleration  vector  A  are  dejined  as  fol- 
lows by  means  of  their  pro- 
jections, denoted  in  the  usual 
manner : 

dv,,      d^y        ,,,^, 
Fig.  20  "       dt        df       "^  ^  ' 

and  if  /^  and  f^  are  the  projections  of  the  force  vector  F,  we  have 

expressing  analytically  the  fundamental  relation  between  accelera- 
tion and  force  vectors.  For  the  magnitudes  of  these  vectors  we 
have 


dx 
dt 


X 


=v^^ 


Y 

d'/ 

/     1 

^^                            i 

dt' 

L- 

^^^\        1 

The  magnitude  of  the  ve- 
locity is  called  the  speed. 
The  essential  distinction 
between  speed,  which  is  a 
positive  number,  and  veloc- 
ity, which  is  a  vector,  is 
one  frequently  overlooked, 
and  serious  errors  result. 
Finally,  we  have  for  the  direction  of  the  velocity  vector 

*  Note  the  close  analogy  of  these  definitions  with  the  definitions  of  velocitj'  and 
acceleration  in  a  straight  line  (p.  42).  The  only  difference  is  that  vectors  replace 
numbers. 


(  Uj  is  negalive) 
Fig.  21 


dx 
dt- 


PLANE  MOTION  57 

^        cos{x,v)  =  -sin(7j,v)  =  -^ 
tan  (x,  v)  =  ~  ' 

cos  (y,  v)  =  +  sin  {x,  v)  =  — 

and  similarly  for  the  other  vectors. 

19.  Some  geometric  properties  of  plane  motion.  In  considering 
rectilinear  motion  our  first  concern  was  to  fix  a  positive  sense  along 
the  line.  So  in  the  plane  curve,  which  is  now  to  be  our  path,  we 
fix  a  convenient  point  from  which  to  measure  the  arc  s,  and  fix  a 
positive  direction  along  tlie  curve.  We  shall  then  denote  by  t  the 
direction  of  the  tangent  to  the  curve  at  the  point  considered,  the 
positive  sense  of  its  tangent  agreeing  with  the  positive  sense  of 
the  curve.  We  shall  denote  by  n  the  normal  obtained  by  rotat- 
ing the  tangent  through  an  angle  +  7r/2.  The  terms  tangential 
velocity,  normal  velocity,  tangential  acceleration,  and  normal  accel- 
eration will  then  readily  be  understood  as  the  projections  of  the 
vectors  named  upon  the  directions  named.  We  are  now  concerned 
with  these  vectors  and  with  these  projections. 

(a)  The  velocity  vector  is  tangent  to  the  path  (though  it  may  be 
pointed  in  the  negative  sense  of  the  tangent).  For,  from  the  cal- 
culus, T  being  the  angle  between  the  tangent  to  the  curve  and  the 
a;-axis,  tan  t  =  {dy/dt)  -s-  (dx/dt).  But  this  is  v^/v^  =  tan  (x,  v),  so 
that  tan  r  =  tan  (x,  v)  and  the  theorem  is  true. 

(b)  Tangential  and  normal  velocities.  From  the  above  it  is  at 
once  evident  that  the  normal  velocity  is  zero.  For  the  tangential 
velocity  we  have,  referring  to  IV,  §  5,  t',=  'y^cos(^,  ^)-l-  VyCos(y,  t). 
But  in  tlie  calculus  it  is  shown  that 

dx      dx      ds 

cos  (X,  t)  =  cos  T  =  -—  =  -;-  -i-  —  » 

^  '   '  ds      dt      dt 

/     ^\      dy      ds 

Using  these  values  with  v.^  =  dx/dt   and  v^  =  dy/dt,  we  obtain 


58 


DYNAMICS  OF  A  PARTICLE 


\n 


y^- 


Vf  =  ds/dt;  that  is,,  the  tangential  velocity  is  the  derivative  of  the 

arc  with  respect  to  the  time* 

(c)  The  acceleratio7i  vector  is  not,  in  general,  tangent  to  the  path. 

For  this  would  mean  ayl a,^  =  vjv^,  or  {dVy/dt) -^ (dvjdt)  =  v,Jv^; 

i.e.  dVy/Vj^  =  dv^/v^,  and  integrating,  log  v^  =  log'y^+logw,  whence 

VyZ=  mv^  or  dy/dt ^iin^dx/dt), 
and  integrating  again,  y  =  mx 
+  c.  We  therefore  have  the 
theorem :  If  the  acceleration 
vector  is  continually  tangent 
to  the  path,  the  path  is 
straight.  As  a  corollary  we 
see  that  where  the  path  is 
curved  the  acceleration  is 
not  tangent  to  the  path. 

(d)    The   ta7igential   and 
normal  accelerations.   These, 

like  the  tangential  velocity,  we  obtain  by  application  of  IV,  §  5. 


a„\ 


-L 


(ttn  is  negative) 
Fig.  22 


First 


a^  =  a^  cos  [x,  t)  +  a^  cos  {y,  t) 


dx  dy 

=  a f-  ft^  -^  = 

""  ds        ■'  ds 


(A/tAy       Cv    tAj 


dx  dy 

a \-  a  -^ 

""dt        'dt 


ds 
It 


dy  d'^y\  _  ds 
dt    dt"")  '  dt~~' 

\2dt 

-/dx\'     /dyV- 
\dt)      \dt)  _ 

r 

.  ds 

'  dt 

ds      d^s 

'  'dt~  dt'^' 

dt  de'^ 

~  [2  dt  [dt 

that  is,  the  tangential  acceleration  is  the  second  derivative  of  the  arc 
with  respect  to  the  time.    For  the  normal  acceleration 

a^  =  a^  cos  {x,  n)  +  a^  cos  {y,  n)  =  —  ct^  cos  {y,  t)  +  ci^  cos  {x,  t) 
dy 
ds 


=  —  a. 


dx      I     dx  dy\  dt 


ds 


*  Notice  that  the  tangential  velocity  may  be  positive  or  negative  according  as  the 
point  is  moving  in  the  positive  or  negative  direction  along  the  curve.  The  speed  is 
the  (positive)  numerical  value  of  the  tangential  velocity. 


PLAKE  MOTION  59 


To  obtain  an  interpretation  for  tliis  we  refer  again  to  the  calculus, 
where  we  find  for  the  radius  of  curvature  the  expression 

dt. 


B  = 


'■  ,  /dyy]^     (d\  dx      d^x   dy\      /dsV     , 

whence  a^v^  —  a^v^  =  l-£\^B. 

We  have,  therefore,  a„  =  (-rl  -^ B  —  —  > 

"     \dt/  B 

or,  the  normal  acceleration  is  the  square  of  the  speed  divided  hy  the 
radius  of  curvature. 

The  usual  formula  for  the  radius  of  curvature  is 


-[-(l)T-f. 


and  the  values  _iZ  =  _^  _; 

dx       dt       dt 

dx^~\dt\dxj     ■  'dt~\dt'dfi      'dt'dt^j  '  \dt) 

reduce  this  to  the  form  given. 

In  the  numerator  of  R  occurs  a  square  foot,  and  the  reduction  made  in 

deriving  the  expression  for  a„  amounts  to  giving  to  this  square  root  the 

sign  of  ds/dt.    The  result  is  that  R  must  be  considered  positive  or  negative 

according  as  the  center  of  curvature  lies  in  the  positive  direction  of  the 

normal  to  the  curve  as  above  fixed,  or  in  the  negative  direction.    The 

formtilas  found  in  (b)  and  (d)  are  important,  and  are  therefore  gathered 

together  here  :  ,  ,00 

°  ds  n  d^s  ir 

„,  =  _,     „„  =  0,     a,  =  ^,>     a.  =  - 

The  two  projections  of  the  acceleration  are  interesting  in  connection  vrith 
their  effect.  The  formulas  show  that  the  effect  of  the  tangential  accelera- 
tion is  exclusively  to  change  the  tangential  velocity  ;  the  effect  of  the 
normal  acceleration  is  exclusively  to  change  the  curvature  of  the  path.  It 
is  easily  seen  from  physical  considerations  that  the  normal  component  of 
the  acceleration  points  from  the  curve  toward  the  center  of  curvature. 

IX.   PROBLEMS  ON  PLANE  MOTION 

In  the  following  problems  derive,  when  possible,  the  equation  of  the 
path  by  eliminating  t  from  the  parameter  equations  ;  find  also  expressions 
for  v„  Vy  and  a^,  a^  Plot  the  path,  and  draw  in  different-colored  inks  for 
various  values  of  t  the  velocity  vector  (red,  say)  and  the  acceleration  vector 


60  DYNAMICS  OF  A  PARTICLE 

(blue,  say),  and  the  tangential  and  normal  components  of  the  latter  (blue 
dotted,  say).  Add  a  descri^jtion  of  any  salient  characteristics  of  the  motion. 
A  different  scale  may  be  used  for  position  vector,  velocity  vector,  and 
acceleration  vector  in  case  any  become  too  large  or  too  small  to  yield  a 
clear  figure,  but  in  this  case  the  scales  should  be  clearly  designated. 

1.  x  =  t^\ 

Solution.  The  path  is  seen  to  be  y^  =  x^,  the  "  semicubical  parabola." 
By  differentiation  we  find  i\  =  2 1,  Vy  =  3  t"^,  and  a^  =  2,  a^,  =  6 1.  We  next 
proceed  to  draw  the  path  and  the  components  parallel  with  the  axes  of 
velocity  and  the  total  velocity,  first  making  a  table  (see  Fig.  23).  We 
may  always  compute  the  magnitude  of  the  velocity,  or  the  speed: 


(The  vertical  bars  about  t  here  denote  that  the  positive  numerical  value  of 
Ms  to  be  taken.)  We  see  from  this  that  the  speed  vanishes  for  t  =  0  and 
increases  with  |/|.  As  vy  =  3  t-  is  always  positive  except  for  t  —  0,it  follows 
that  the  motion  is  always  upward. 

We  next  turn  to  a  consideration  of  the  acceleration.    The  components 

may  be  drawn  as  before  for  the  velocity,  being  previously  tabulated  (see 

Fig.  2i).    The  resultant  should  then  be  drawn. and  a  tangent  and  normal. 

The  tangent  should  be  drawn  by  means  of  its  sloj^e,  and  not  merely  by 

f  a  ruler  laid  along  the  curve.  The  tangential  and  normal  accelerations  are 

*  the  projections  on  tangent  and  normal  of  the  acceleration  vector. 

Remark.  Analytic  expressions  for  tangential  and  normal  acceleration 
may  be  calculated  by  the  formulas  given  on  p.  59.  The  only  difficulty  con- 
sists in  determining  the  sign  of  the  radical  in  ds/dt = »t  =  V(dx/dty+ (dy/dty. 
In  our  present  problem  if  we  measure  s,  say,  from  the  origin,  and  always 
increasing  with  y,  then  as  y  always  increases  with  t,  so  also  does  s,  and 
ds/dt  is  positive  or  zero.    Hence 


2.  x  =  /2-|        3.  x  =  M        4.  x  =  l  +  cos/     ^       5.x  =  t  \ 

y=2ty  y  =  t^J  y  =  2cos(t/2)y  y  =  l/(\  +  t^)r 

Take  for  t  in  problem  4  a  series  of  values  0,  7r/3,  2  tt/S,  tt,  •  •  • . 


x  =  t 


^^  8.  x  =  cost    \ 

y  =  2Vl-t^j  y  =  2sint)' 


=  cosA-n  9.  x  =  (l-t^)/(l  +  t^)^ 

=  sinktj'  y  =  2t/(l  +  t")  }' 


I .   X 

y 

The  path  is  closed.    How  long  does  it  take  for  the  body  to  make  a 
complete  circuit  of  its  path? 


PROBLEMS  ON  PLANE  MOTION 


61 


'      1      f 

1    J- 

^                                          t     t 

27      4       -••' 

7 

r 

1 

t 

M4^^ 

~~T^ 

t 

1 

T 

>_ 

LI. 

f,j 

IT 

IJ 

ct- 

7/ 

~i 

7 

/ 

i' 

It    1—                  _i 

1                  /  T 

/ 

~    Y 

\  "      ^^     " 

^    ^^?^t—  -» 

J     „    l\,X  \T 

'^  <   s\     ^ 

^^\     5 

^S    \ 

S^   A 

^     e 

\   ^ 

^e^r- 

*r                     4^      X 

-e               t^i^\ 

^,      V 

^\ 

^Y 

Cv 

W 

a 

^ 

SI 

\ 

5 

1. 

^ 

1^ 

r 

3 

± 

i               At 

^ 

^ 

c 

Fig.  23  (Velocities) 


62 


DYNAMICS  OF  A  PARTICLE 


Fig.  24  (Accelerations) 


SPECIAL  PLANE  MOTIONS  63 


10.  x  =  t       \  11.  x  =  \og(t  +  Vl-{- f^)\ 

ij  =  sin  tj  y  =  Vl  + 1^  J 

Show  that  tlie  path  is  a  catenary  y  =  ^  (e*  +  e-  *)  and  that  it  is  described 
with  constant  speed. 

12.  Show  that  the  normal  acceleration  is  the  product  of  the  tangential 
velocity  by  the  rate  of  change  of  direction  of  motion,  i.e.  a„  =  (^cls/dt)  •  (dr/dt) 
where  t  =  arctan  Vy/v^. 

20.  The  differential  equations  of  plane  motion.  The  differential 
equations  of  plane  motion  were  expressed  in  the  last  paragraph  in 
the  relation  between  force  and  acceleration.  Usually  they  are 
written 

The  problem  of  determining  the  "  complete  description  "  from  these 
equations  together  with  proper  auxiliary  conditions  is  one  whose 
difficulty  depends  entirely  upon  the  nature  of  the  force.  The 
simplest  case  is  that  in  which  the  component  /^  depends  only 
upon  X  and  dx/dt,  and  f^  only  upon  y  and  dy/dt.  In  this  case 
the  solution  of  each  differential  equation  is  a  separate  problem. 
Usually,  however,  both  /^  and  f^  depend  upon  both  x  and  y  and 
upon  v^  and  Vy.  The  equations  are  therefore  simultaneous  differ- 
ential equations,  and  are  solved  by  various  devices  depending 
upon  the  special  problems  attacked,  important  among  which  is 
the  elimination  of  one  of  the  dependent  variables. 

If  we  know  the  patli  in  advance  (see  footnote,  p.  55),  we  only 
need  determine  the  function  s  =  ^{t)  giving  the  way  in  which  the 
path  is  described.  To  this  end  we  use  instead  of  the  differential 
equations  (1)  the  single  equation  which  comes  from  considering 
tangential  components  ('see  §  6) : 

dH      f 
de 

21.  Some  special  plane  motions. 

(a)  Circular  motion.  Let  Q  denote  the  angle  between  the  avaxis 
and  a  line  joining  the  origin  to  the  point  (x,  y),  which  we  suppose 


64 


DYNAMICS  OF  A  PARTICLE 


to  move  on  a  circle  of  radius  r  about  the  origin.  Let  us  count  the 
arc  s  of  this  circular  path  from  the  point  where  it  crosses  the 
avaxis.  Then  x  —  r  cos  0,  y  =  sin  6,  and  s  =  rO.  From  the  pre- 
ceding section  we  then  have,  by  differentiating  both  members  of 
the  equation  s  =  rd. 


and 


cW 
dt 


a_  =  ,       ,^r  =  r 


(Is 


ds 


dh     d'e 

dt)  '  ^~^\dt/' 


The  quantities  o)  =  dd/dt  and  a  =  d^6/dt^  are  called  respectively 
the  angular  velocity  and  the  angtdar  acceleration,  and  are  the  time 
rates  of  change  of  the  angle  and  of  the  angular  velocity  to.  These 
notions  have  also  an  important  application  to  the  study  of  the 
rotation  of  a  rigid  body  about  an  axis  (see  p.  91). 

As  an  illustration  of  the  general  method  for  obtaining  projections  upon 
given  lines  of  the  velocitj'-  and  acceleration  vectors,  we  shall  derive  these 
results  directly,  obtaining  incidentally  an  interesting  verification  of  our 
general  theory. 

Using  IV  of  §  5,  we  have  t",  =  v^co^{x,  t)  +  t>j,cos(?/,  t).  But  as  (x,  t)  = 
6  +  Tr/2  and  (jj,t)  =  6,  cos  (z,  <)  =  —  sin  6  and  cos  (y,t)  =  cos  6 ;  moreover 

dx 


Y 

11^ 

\ 

0 

\ 

X 

and 


And  — 
lit  fit 

(hi         ^iie 

-^  =  r  cos  a 

dt  dt 


Using  these  values,  we  have 

V,  =  r(s\n-d  +  cos^^)  —  =  r  — 
"^  dt         dt 

Again,  by  differentiating  v^  and 
Ave  have 


=  —  r  cos  ^  (  ^  I  —  r  sin 


Fig.  25 


le 

dt 
16V 


,^m 


df^j' 

Forming  now  a,  after  the  manner  of  r„  we  have  at  =  r(d-6/dt'^);  and 
forming  a„,  keeping  in  mind  that  (x,  n)  =  6  +  ir,  (ij,  n)  =  6  +  Tr/2,  so  that 
cos(x,  n)  =  — cos^,  cos(?/,  n)  =  - sin^,  we  have  a„  =  r(dO/dty,  all  these 
values  agreeing  with  those  found  above.  The  results  may  be  stated :  The 
tangential  velocity  is  r  times  the  angular  velocity;  the  tangential  acceleration  is  r 


PROJECTILES 


65 


times  (he  angular  acceleration ;  and  the  normal  acceleration  is  r  times  the  square 
of  the  angular  velocity. 

Ex.  1.  If  0)  is  constant,  6  =  <ot.  Using  the  equations  z  =  r  cos  wt,  y  — 
rsinwt,  find  directly  the  values  for  v^  and  a„,  and  show  that  a^  =  0.  Show 
that  the  force  producing  the  motion  is  always  toward  the  center  (called 
centripetal  force).  The  point  on  the  ar-axis  given  by  x^rcostat  is  the 
projection  of  the  moving  point.  Its  motion  is  simple  harmonic  motion 
(see  p.  48). 

(b)  Projectiles.  If  a. body  is  thrown  into  the  air  and  left  to  the 
action  of  gravity  alone,  resistance  of  the  air  and  the  effect  of  the 
rotation  of  the  earth  being  neglected,  the  differential  equations 
become 


cie 


0 


df  -^ 


y, 


d^:r. 

] 

0 

df 

dSj_ 

9 

de 

If  we  take  for  origin  of  coordinates  the  point  from  which  the  body 
was  thrown,  and  if  the  body  was  thrown  in  a  direction  making  an 


Va^in  7", 


/  VoCOS  T, 


Fig.  26 

angle  r^  with  the  horizontal  with  an  initial  speed  Vq,  the  auxiliary 
conditions  are  «  =  0,  y  =  0 ;  v^=  v^  cos  Tq,  v,j  =  v^  sin  Tq  for  ^  =  0.  A 
first  integration  of  tlie  differential  equations  gives  dx/dt,  or  v^=  c^; 
dy/dt,  or  Vy  =  —  gt  +  c^.  When  ^  =  0  these  reduce  to  %  cos  t^  and 
Vq  sin  Tq  respectively.    Hence  c^  =  %  cos  r^  and  c^  =  %  sin  t,,.   Hence 


66  DYNAMICS  OF  A  PAETICLE 

!  ■■ 
\l-y,       dx/dt  =  Vq  cos  Tq  and  dy/dt  =  —  gt  +  v^  sin  t^.    Integrating  again, 

we  have 

X  =  (Vo  cos  T^)t  +  c^,     y  =—\ gf"  +  (%sin  r^) t  +  c,. 

As  re  =  0  and  y  =  Q  where  (5  =  0,  Cj  =  0  and  c^  =  0.    Hence  we  have 
the  complete  description 


X  =  (VoCOSTo)^ 


r,)t]' 


Some  important  results  in  the  theory  of  projectiles  are  now  left  for  the 
student  to  derive  in  the  ensuing  exercises  and  in  some  later  problems. 

Ex.  2.  Verify,  by  eliminating  t,  that  the  equation  of  the  path  of  the 
projectile  is 

y  =  „,  ~^^    ,,  +  (tan To) X. 

2  {Vo  cos  To)2 

What  is  the  nature  of  this  curve?  Draw  it  for  tq  =  45°,  Vq  =  40  ft.  per 
sec,  using  g  =  32.    Draw  the  velocity  vectors  for  t  =  0,  t  =  ^,  t  =  1,  t  =  ^. 
Ex.  3.  Range.  By  the  range,  R,  oi  a  projectile  is  meant  the  horizontal 
distance  covered  before  it  reaches  the  ground  again.    Show  that 

^_i£_sin2_To_ 

g 

What  elevation  gives  the  greatest  range?  Show  that  any  two  values  of 
To  differing  by  equal  amounts  on  either  side  of  this  elevation  give  the  same 
range. 

Ex.  4.  Find  the  time  of  Jlight.  The  initial  speed  i'q  being  given,  what 
elevation  gives  the  greatest  time  of  flight  ? 

Ex.  5.  Suppose  the  initial  speed  t'o  be  given.  Then  for  various  values 
of  To  we  have  a  system  of  parabolas  with  vertical  axes  and  all  passing 
through  the  same  point.  Any  point  in  the  plane  which  is  covered  by  this 
system  is  tvitkin  range,  that  is,  can  be  hit  with  the  given  initial  velocity. 
In  other  words,  all  points  within  the  envelope  of  the  system  are  within 
range.  Find  the  envelope,  writing  the  system  in  the  form  y  =  — 
gx^{l  +  m^)/2v^  + mx.  The  result  should  be  a  parabola  whose  highest 
point  is  that  of  the  highest  path  of  the  system  and  whose  breadth  at  the 
level  of  the  initial  point  is  twice  the  maximum  range.  Show  that  your 
result  agrees  with  these  statements. 

Ex.  6.  Refer  the  path  of  the  projectile  of  Ex.  2  to  its  vertex  by  a  shift 
of  the  axes  (see  Analytic  Geometry).  What  are  the  coordinates  of  the 
vertex?  What  is  the  latus  rectum?  Show  that  all  the  trajectories  corre- 
sponding to  a  given  initial  speed  have  the  same  directrix. 

(c)  Pendulum.  We  are  dealing  with  a  particle  moving  in  a 
circular  arc  under  gravity.    Knowing  the  path,  we  proceed  as 


PENDULUM 


67 


indicated  in  §  20,  except  that  instead  of  s  it  will  be  convenient  to 
use  the  angle  between  the  pendulum  and  a  line  vertically  down- 
ward. Then  s  =  W,  I  denoting 
the  length  of  the  pendulum ; 
and  as  the  tangential  compo- 
nent of  the  force  is  —  m(/  sin  d, 
we  have 

d^s         jd^9  .    ^ 

m  — ^  =  ml  -— r  =  —  mq  sm  a, 


or 


df 
d'd_      g 


de 


df 


=  -^sin6'. 


To  avoid  difficulties  in  integra- 
tion it  is  usual  to  substitute  for 
sin  6  on  the  right  the  quantity 
6,  which  approximation  *  is  in 
error  by  less  than  ^V6,  where 
6  is  measured  in  radians,  and 
which  is  quite  satisfactory  when 
6  does  not  exceed  a  twentieth 
of  a  radian.  The  resulting  equa- 
tion,  ^^ 

de~ 


Fig.  27 


I 


e, 


is  easily  integrated  (see  (b),  §  16),  and  yields 


=  ^lsin(  x||-  0+-^cos 


*  If  this  approximation  is  not  used,  the  method  is  as  follows:  Multiplying  l)y  cW/dt 
(see  (c),  §  10),  the  equation  hecomes 

d  l/ddY     g    d        ^        ,  /(?9\2     2g 

dt~2[dt)=l-dt'''''^'    ^'^^^"''^     UJ=T' 
Let  the  pendulum  start  from  an  angle  Oq,  so  that  d9/dt  =  0  when  6=  0q;  then  Cx  = 
—  (2  g/l)  cos  ^o»  and  we  have 


C0S^+  Ci- 


(r 


2o 

~j-  (cos  6  —  cos  ^o)    and 


V¥'  +  '-/: 


dd 


'  "^  V cos  0  —  cos  ^0 

This  integral  cannot  be  expressed  by  means  of  elementary  functions,  and  integration 
by  series  or  by  means  of  elliptic  integrals  must  be  resorted  to.  In  order  to  identify 
this  with  an  elliptic  integral,  we  must  reduce  it  to  a  standard  form  by  substitutions 
(see  B.  O.  Peirce,  A  Short  Table  of  Integrals).   First  write  cos^=  1  —  2siu2(^/2), 


68 


DYNAMICS  OF  A  PARTICLE 


If  we  count  the  time  so  that  ^  =  0  when  ^  =  0,  we  find  ^  =  0,  so 
that  the  solution  becomes 

^  =  ^  sin /-J|.  A 

The  motion  is  thus  oscillatory.  The  period,  2  ir^l/g,  or  the  time 
of  a  complete  swing  to  and  fro,  is  seen  to  be  independent  of  A,  i.e. 
of  the  length  of  the  swing.  As  T  and  /  are  observable,  the  pendu- 
lum gives  a  useful  method  of  determining  the  value  of  g.  The 
motion  in  the  circular  arc  is  simple  harmonic  for 


s  = 


e     A  . 


(if 


(d)  Motion  tender  central  forces.  By  a  central  force  is  usually 
understood  one  whose  direction  is  always  in  the  line  joining  the 
particle  with  a  fixed  point,  or  center,  and  whose  magnitude  depends 

only  upon  the  distance 
of  the  particle  from  this 
center.  We  shall  con- 
sider first  a  special  case, 
namely  that  in  which 
the  force  is  always  di- 
rected toward  the  cen- 
ter and  whose  magni- 
tude varies  with  the 
distance  from  it.  Then 
f  =  mJi^r,  where    r  = 


T 

-ff 

^  (^,  y) 
1 

J>^r 

,--*'^ 

t 

,-'  r 

1 

1 
1 
1 
1 

0 


X 
Pig. 


28 


cos^o  =  1  —  sin2  {Oq/2)  ;  then,  introducin 
8in(^o/2)sin0,  we  obtain 
dd 


f 


VcosO  —  COS&o 


!)=». 


a  -new  variable  </>,  by  the  relation  sin  0/2  ■. 
:  =  V2/'(<^,8in|)-         - 

It  t  =  0  when  ^  =  0,  and  hence  when  ^  =  0,  we  have,  since  ^(o, 

,  =  ^f(*,»ta«-»). 

When  6=  0(^,(p=  Tr/2,  so  that  the  whole  time  of  an  oscillation  T,  being  four  times  the 
time  consumed  as  0  goes  from  0  to  0q,  is  4  (l/g)  F[{-ir/2),  sin  (^o  /2)].  As  an  illustra- 
tion the  student  may  show  that  the  period  ^rf_a  pendulum  swinging  through  an  angle 
60°  to  either  side  of  the  vertical  is  6.744  'vl/g.  A  further  study  of  the  properties  of 
elliptic  integrals  is  recommended  at  this  point  to  students  who  find  the  subject  inter- 
esting (see  Byerly,  Integral  Calculus). 


CENTRAL  FORCES 


69 


"saf+y^y  the  force  center  being  taken  as  origin,  and  cos {x,  F)  = 
—  x/r,  cos  {y,  F)  =  —  y/r,  so  that  fx  =  —  mlc'x  and  fy  =  —  mBy. 
The  dififerential  equations  of  motion  therefore  take  the  form 


d^x 
df 

-l^x 

dhj 
df 

- 

-h'y 

J 

Referring  to  (b),  §  17,  we  see  that  the  projections  of  the  motion 
on  the  axes  are  both  simple  harmonic  motion.  To  find  the  path, 
we  endeavor  to  eliminate  t  from  the  solution 

x  =  c^  cos  kt  +  ^2  sin  kt 

y  =  c^  cos  kt  +  c^  sin  kt 

This  is  best  done  by  solving  for  sin  kt  and  cos  kt  and  squaring  and 

adding.    We  find  * 

7 .      c*'^  —  CoV         .    7  ^      —  c^x  +  c.y 
cos  kt  =  -^ ^  ,     sni  kt  = ^ ^  , 

whence  {c^x  —  c„yY  +  (—  c^x  +  c^yY  =  {c^c^  —  c^c^"^. 

This  is  a  conic  section  referred  to  its  center,  for  it  is  of  the  second 

degree  and  the  linear  terms  are  absent.    That  it  is  an  ellipse  is 

easily  verified  by  the  test  given  in  Analytic  Geometry,  namely 

that  B^—  ■iAC<  0  where  A,  B,  and  C  are  the  coefficients  of  a^,  xy, 

and  y^  respectively.    The  motion  is  therefore  called  elliptic  har- 

mo7iic  motion. 

Ex.  7.  Show  that  the  force  producing  the  motion  x  =  a  cos  I't,  y  =  b  sin  kt, 
is  toward  the  origin  and  varies  with  the  distance  from  the  origin. 

Returning  to  the  general  problem  of  central  motion,  we  shall 
have  the  same  direction  cosines  for  the  force,  but  the  magnitude 
/  will  be  some  function  of  r,  say,/=  ir{r).  The  equations  become 

d'x       ,  ,     x' 

-de^^-^^'^-r 
^^V    .» u.^  y 


m 


df 


*  Unless  C1C4  —  C2C3  =  0.    The  motion  in  this  case  is  simple  harmonic  motion  in  a 
straight  line,  as  the  student  may  show  as  an  exercise. 


70 


DYNAMICS  OF  A  PARTICLE 


These  equations  present  the  difficulty  mentioned  in  §  20,  that 
both  equations  contain  both  variables,  since  r  —  Va;"  +  y^.  Two 
integrals  may,  however,  be  obtained, —  first,  the  integral  of  areas. 
Multiplying  the  first  equation  by  —  y  and  the  second  by  +  oc  and 
adding,  we  have 


cl^x         d'-i/ 

^  (If      cie 


that  is. 


dt 


dt 


t^-f-  —  1/  -f^  ]=  0,     or     x-~-~y 


dt 


dx 
dt 


The  meaning  of  this  is  clearer  upon  using  polar  coordinates,  when 
it  becomes 

r  cos  Ol  —  sin  6  +  r  cos  6  —  )  —  r  sin  ^  (  —  cos  ^  —  r  sin  ^  —  I  =  c,, 
\dt  dt  \dt  dt         ' 


or 


r'dB 
dt 


=  Cv 


Now  let  A  represent  the  area  swept  out  by  the  moving  radius 
vector.    We  have,  by  the  calculus, 

dd      2  dA 


A  =  \  CrMd,     or     dA  =  1  rMd,     so  that     —  - 
^J  "  dt 


dt 


and  the  equation  becomes 
dA      Cj 
'dt^2' 


so  that     A  =  —  t  +  c„ 
2  ' 


If  we  count  the  area  from  the 

time  ^  =  0,  we  have  c,  =  0,  and 

SY^^'^P       we  may  say  the  area  swept  out 

t-^i\  hi/  the  radius  vector  in  the  case 

of  a  particle  moving  binder  a 

f ^=0     central  force  is  proportional  to 

the  time  consumed. 

The  second  integral,  known 
as  the  energy  integral,  results 
from  observing  that  cr/dx  = 
x/r  and  cr/dy  =  y/r,  as  may  be  seen  by  partial  differentiation  of 
the  equation  ?-^  =  yf+  y'^.    The  equations  of  motion  thus  become 


Fig.  29 


PLANETARY  MOTION 


71 


dt-       ^  ^  'XcxJ  ex     ^  ' 

^-4  =  -^  (r)(—    = ^rir) 

df-       ^^\dy)         cy     ^y 

where  "^{r)  is  the  negative  of  the  integral  mth  respect  to  r 
•^(r).    Multiplying  by  dx/dt  and  dy/dt  and  adding,  we  have 

dxd'^x  .    dy  d^y\ d'^{x)    dx      d'^{r)    dy 

dt  dt'^  j  ex 

d^    mv^  _      d^jr) 
dt'         ~ 


of 


dt  dt^ 


dt 


cy 


dt 


or 


~2~ 


dt 


whence 


The  first  term,  |-  miy^,  is  the  Jcinetic  energy  (see  §  2  5)  of  the  particle, 
and  the  second,  "^(r),  is  called  the  potential  energy  (see  §  24);  the 
equation  states  that  the  sum  of  the  kinetic  and  p)otential  energies  of 
the  particle  is  constant.  Tliis  is  a  special  case  of  an  important  prin- 
ciple in  mathematical  physics,  known  as  the  conservation  of  energy. 
We  consider  now  briefly  the  motion  of  the  planets  about  the 
sun.  If  we  consider  these  bodies  as  spheres,  the  forces  they  exert 
on  each  other  may  be  regarded  as  acting  at  their  centers,  as  is 
shown  in  the  theory  of  attraction.  Let  us  take  the  sun  at  the 
origin  and  consider  one  of  the  planets.  It  will  be  attracted  (see 
(c),  §  17)  with  a  central  force  whose  magnitude  is  proportional  to 
the  reciprocal  of  the  square  of  its  distance  awayj  say,  k/r^.  The 
force  being  attractive,  we  have 


T/r(r)  =  — -     and     "^(r) 


'*fi 


dr=^-- 


The  force  exerted  by  the  other  planets  is  small  in  comparison  with 
that  exerted  by  the  sun,  so  we  shall  get  an  approximation  to  the 
motion  if  we  neglect  it.  The  energy  integral  becomes  ^  mv^  —  k/r 
=  Cg,  or,  using  polar  coordinates,  since 

dt. 


(ds\ 

~\dtj 

2 

\dti  ^ 

\Jt) 

2 
+ 

\dt] 

72 


DYNAMICS  OF  A  PARTICLE 


We  may  now  obtain  the  differential  equation  of  the  path  if  we 
can  eliminate  t.  This  may  be  done  by  the  integral  of  areas,  which 
may  be  written  dO/dt  =  cjr^.  We  have  then,  since  dr/dt  = 
(drJdB)  ■  {dd/dt), 


ih" 


dd\ 

dt) 


=n(rj-'](f)-^- 


1 

-  m 
2 


so  that  the  differential  equation  becomes 

This,  solved  for  dr/dd,  yields 
dr 

Td 


=  c,. 


Whence,  integrating,  0  =   i  = 


dr 


2  k 
H i^  — 1 


2  k 

+  — i^-1- 
met 


+  c,. 


This  integral  may  be  evaluated  by  a  substitution  r  =  \/u,  or  by 
reference  to  tables  (e.g.  Peirce's  Short  Table  of  Integrals,  No.  183), 
which  gives 


/      — -^r—  2 


6  —  c^=  arcsin 


k~ 


me: 


\^k:' 
r  \\ 


=  arcsin 


+ 


8  c, 


\-\/Tc^+  2mc^c^ 


mc,' 


,       mc' 

k 

r 


This  may  be  written 

sin(^  — c,)= 

y/l^  +  2  7nc^c^ 

or  if  we  suppose  c^  =  — 7r/2,  which  amounts  only  to  a  particular 
choice  of  the  direction  of  the  a:-axis,  we  have 

sin  (^  —  c^)  =  sin  I  ^  +  -  I  =  cos  0, 

and  the  equation  may  be  written 


mc{ 


=  1- 


>^'l  +  ^^C08^. 


PROBLEMS  ON  PLANE  MOTION  73 

Comparing  this  with  the  equation  (see  Analytic  Geometry)  I  /r  == 
1  —  e  cos  6,  we  see  that  the  path  is  a  conic  section  referred  to  its 
focus  as  origin.  This  fact  was  discovered  by  observation  by  Kepler, 
and  bears  the  name  Kepler's  first  law.    The  eccentricity  is 


=V 


1  I  2mc^g3 


and  is  =1,  according  as  2  mc^c^/k=.0,  that  is,  according  as  CglO. 
But,  referring  to  th6  energy  integral,  we  find  an  interpretation 
for  Cg.  For  if  we  denote  by  v^  and  r^  the  speed  of  the  planet  and 
its  distance  from  the  sun  when  ^  =  0,  we  have,  since  the  energy 
equation  holds  for  every  t,  c^=  ^ mv^—{h/r^.  Hence  e%l  accord- 
ing as  Cg  1 0,  that  is,  according  as  v^  1 2  h/mr^.  Thus  the  path  will 
be  a  parabola  if  the  initial  speed  is  V2  h/mr^,  and  this  no  matter 
what  the  initial  direction  of  the  planet.  For  larger  initial  speeds 
the  paths  are  hyperbolas,  for  smaller  ones  ellipses.  Planets  move 
only  in  ellipses.  Comets  have  orbits  of  any  of  the  types.  The 
earth's  orbit  is  very  nearly  a  circle,  its  eccentricity  is  about  J-^. 

The  motion  of  the  body  in  its  path  is  determined  by  the  inte- 
gral of  areas.  Equal  areas  being  swept  out  in  equal  times,  it  follows 
that  the  planets  move  fastest  when  nearest  the  sun.  This  result 
may  also  be  read  from  the  energy  integral 


X.  PROBLEMS  ON  BODIES  MOVING  IN  A  PLANE  UNDER  THE  ACTION 

OF  FORCES 

1.  A  snowball  fight  is  arranged  between  two  boys  who  can  throw  with 
speeds  of  78  ft.  per  sec.  and  83  ft.  per  sec.  resjiectively.  In  order  to  equal- 
ize the  contest  the  boy  with  the  less  speed  is  given  a  station  6|  ft.  higher 
than  that  of  his  comrade.  Assuming  that  the  effect  of  the  snowball  varies 
with  tlie  square  of  the  speed,  does  this  arrangement  equalize  the  figlit? 

Solution.  We  must  first  find  the  expression  for  the  speed  of  a  projectile. 
Taking  the  equations  of  motion,  finding  v.j^  and  Vy,  and  squaring  and  adding, 
we  have  v^  =  v^  —  2  gt  (jjq  sin  Tq)  +  g^fi-  We  wish  to  know  the  value  of  this 
expression  for  a  given  value  of  y.  To  do  this  it  is  only  necessary  to  find  t 
from  the  equation  for  y,  y  =  —  fffi/2  +  (wosin  To)<,  and  substitute  it  in  the 
equation  for  v^,  or,  in  other  words,  to  eliminate  t.  This  is  easily  done  by 
multiplying  the  second  equation  by  2  g  and  adding.  The  result  is  v^  +  2  gy  = 
„2  _  „2  _  2  gy.    Now  let  v'  be  the  sjieed  with  which  the  stronger  boy 


74  DYNAMICS  OF  A  PARTICLE 

hits  the  weaker  at  a  height  6{  ft.  above  him  ;  then  v'^  =  (83)2  _  2  «7(6i)  ; 
and  if  v"  be  the  speed  "with  which  the  weaker  boy  hits  the  stronger  at  a 
height  6{  ft.  below  him,  then  v"^  =  (7S)^  +  2 g{6\).  Computing  these 
with  g  =  32.2,  we  find  v"^  =  v"-  =  6486.5,  that  is,  the  handicap  was  correct. 
Remark.  The  striking  circumstance  in  this  jiroblem  is  that  in  the  elimi- 
nation of  t  between  the  expressions  for  y  and  v^,  tq  drops  out,  and  we  find 
that  with  a  given  initial  velocity  the  speed  depends  only  on  the  height. 
This  fact  greatly  simplifies  the  solution. 

2.  If  a  man  can  throw  a  baseball  350  ft.,  with  what  speed  does  it  leave 
his  hand? 

3.  Suppose  a  rifle  ball  in  traveling  100  yd.  deviates  from  a  horizontal 
line  by  less  than  1  in.  Show  that  its  initial  speed  must  exceed  2070  ft. 
per  sec. 

4.  What  elevation  will  give  a  range  of  5000  yd.  with  an  initial  speed 
of  1500  ft.  per  sec.  ? 

5.  Show  that  the  heights  to  which  a  body  will  rise  when  projected  with 
a  given  initial  speed,  but  with  elevations  80°,  45°,  60°,  and  90°,  will  be  in 
the  ratios  1:2:3:4. 

6.  A  man  who  can  make  a  standing  broad  jump  of  8  ft.  stands  at  the 
window  of  a  burning  building  60  ft.  above  the  level  of  a  river  whose 
depth  at  any  point  is  one  tenth  its  distance  from  the  face  of  the  building. 
How  deep  water  can  he  reach  by  jimiping?  If  he  dived,  at  what  angle 
would  he  strike  the  water  ?  (Solve  this  problem  on  the  hypothesis  that  he 
jumj^s  into  the  air  at  an  angle  45°.  As  a  matter  of  fact  an  angle  ^  arccos  l|- 
would  be  more  advantageous  in  both  respects,  as  may  be  shown  by  the 
student  who  is  sufficiently  interested.) 

7.  What  area  of  the  surface  of  a  building  can  the  stream  from  a  fire 
hose  cover,  the  nozzle  being  3  ni.  from  the  building  and  the  speed  of  the 
stream  being  20  m.  per  sec.  ?  (See  Ex.  5  under  Projectiles,  §  21.) 

8.  Show  that  the  point  (x^,  y^)  is  within  range  if  y■^^  +  ^/x^+yf  =  v^/g. 
Hint.  Use  Ex.  5. 

9.  Show  that  the  area  of  that  part  of  a  level  plane  which  is  within 
range  of  a  gun  a  height  h  above  the  plane  increases  linearly  with  h,  and 
has  the  form  A  -^  2h  VttA  where  A  is  the  area  within  range  when  the  gun 
is  in  the  plane.  Show  that  the  area  within  range  varies  as  a  biquadratic 
function  of  the  initial  speed. 

10.  Show  that  the  speed  of  a  projectile  is  the  same  at  any  two  points 
of  the  same  height.  Show  that  the  initial  speed  is  the  speed  that  a  body 
would  acquire  in  falling  from  rest  at  the  directrix  of  the  parabola  to  the 
level  of  the  initial  jioint.  Extend  the  reasoning  so  as  to  prove  a  similar 
theorem  for  the  speed  at  any  point. 

11.  A  gun  is  at  the  bottom  of  a  hill  the  angle  of  inclination  of  whose 
face  is  i.    Show  that  the  range  for  an  elevation  Tq  is 


PROBLEMS  ON  PLANE  MOTION  76 

j^_2  v^   cos  Tq  sin  (tq  -  i) 
g  cosH 

that  the  best  range  is  obtained  when  the  initial  velocity  vector  bisects  the 
angle  between  the  face  of  the  hill  and  the  vertical,  and  that  this  best  range 
is,  R-  v^/(j(l  +°'sini). 

12.  The  focus  of  the  trajectory  of  a  projectile  is  above  or  below  the 
horizontal  plane  from  which  it  was  projected  according  as  the  angle  of 
elevation  is  greater  or  less  than  the  elevation  giving  the  maximum  range. 

13.  Water  jets  are  sent  out  in  all  directions  with  a  speed  of  30  ft.  per 
sec.  from  a  fountain,  the  nozzle  being  in  the  form  of  a  pierced  ball.  What 
is  the  nature  and  the  equation  of  the  surface  of  the  water  ? 

14.  Show  that  when  a  number  of  projectiles  are  fired  simultaneously 
from  a  point  in  directions  which  all  lie  in  the  same  plane,  the  projectiles 
will,  at  any  instant,  all  lie  in  a  plane  parallel  to  the  plane  of  the  initial 
velocities.  Resolve  the  acting  forces  and  initial  velocities  into  components 
normal  to  and  parallel  with  the  plane.  Show  that  the  bodies  will  all  move 
the  same  distance  normal  to  the  plane. 

15.  Show  that  if  three  bodies  are  projected  simultaneously  from  the 
same  point  and  in  the  same  vertical  plane,  the  triangle  subtended  by  them 
remains  similar  to  itself  and  has  an  area  varying  with  the  square  of  the 
time. 

16.  A  clock  gains  3  min.  per  day.  What  proportion  of  length  should  the 
pendulum  be  lengthened  or  shortened  ? 

Use  the  results  of  the  approximation  for  small  arcs. 

17.  A  pendulum  making  2  n  swings  per  day  is  lengthened  from  I  to 
I  +  M.    Show  that  it  will  lose  approximately  nAl/'2l  swings  per  day. 

1 8.  In  the  equation  for  pendulum  motion  (cb/dty  =  w^  =  (2  g/l)  cos  6  ■\-  c 
determine  c  by  the  data  w  =  cdq  when  ^  =  0.  Then  find  what  the  initial 
velocity  Wq  must  be  in  order  that  the  pendulum  (considered  as  rigid)  rise 
to  the  vertical  and  continue  revolving,  rather  than  oscillate. 

19.  An  automobile  weighing  1000  lb.  turns  a  corner  on  the  arc  of  a  circle 
of  radius  50  ft.  at  a  speed  of  6  mi.  per  hr.  Find  the  "  skid  "-producing 
force.    In  what  ratio  would  this  force  be  reduced  if  the  speed  were  halved? 

20.  A  centripetal  force  of  131  oz.  holds  a  body  in  a  circle  of  radius 
100  ft.  with  a  uniform  velocity  of  one  revolution  per  hr.  Find  the  weight 
of  the  body. 

21.  Formerly  railroad  curves  were  built  in  the  fonn  of  a  circle  touched 
by  two  straight  lines.  A  later  method  consisted  in  using  hyperbolic  arcs. 
Explain  how  this  was  an  improvement. 

22.  At  what  angle  must  the  track  be  raised  at  a  point  whei-e  its  radius 
of  curvature  is  400  ft.  in  order  that  for  trains  with  a  speed  of  40  mi.  per 
hr.  gravity  may  exactly  balance  the  pressure  on  the  outer  rail  due  to  the 
normal  acceleration? 


76  DYNAMICS  OF  A  PARTICLE 

23.  What  proportion  of  its  weight  does  a  body  lose  at  the  equator 
because  of  the  earth's  rotation  ?  How  many  times  as  fast  would  the  earth 
have  to  rotate  in  order  that  bodies  should  have  no  weight  at  the  equator? 

24.  A  hammer  thrower  whirls  a  hammer,  of  weight  16  lb.,  in  a  circle 
of  radius  4  ft.,  and  the  pull  on  his  arms  just  before  letting  go  is  180  lb. 
Find  the  speed  as  he  lets  go,  and  the  length  of  his  throw,  assuming  the 
hammer  to  rise  initially  at  an  angle  of  45°. 

25.  A  body  moves  with  uniform  speed  in  the  parabola  y^—  \  ax.  Deter- 
mine the  normal  component  of  the  force.  Show  that  it  is  greatest  at  the 
vertex. 

26.  Show  that  if  the  velocity  vector  always  has  the  same  direction,  the 
motion  will  be  rectilinear. 

27.  In  the  elliptic  harmonic  motion  of  the  present  section  determine  the 
constants  of  integration  by  the  following  data  :  when  i  =  0  the  body  is 
moving  directly  upward  from  the  point  (a,  0)  with  a  speed  kb.  Find  the 
equation  of  the  path. 

28.  On  buoyant  bodies  the  upward  pressure  of  the  air  might  be  taken 
into  account  by  assuming  for  the  force  ./^=0,  f,^  =  — mg  +  mk(h  —  y), 
where  A  is  a  h'eight  where  the  upward  pressure  practically  ceases.  Study 
the  motion.    Let  the  body  rise  from  rest. 

29.  Study  the  motion  of  a  jioint  repelled  from  a  fixed  center  by  a  force 
proportional  to  its  distance  from  the  center.  Show  that  the  path  is  a 
hyperbola.  Compare  with  elliptic  harmonic  motion,  and  also,  as  in  prob- 
lem 27  of  this  section,  work  out  a  special  case  with  simple  initial  conditions. 

30.  Study  the  motions  obtained  from  a;  =  «  cos  int,  y  —  h  sin  nt,  by  giving 
m  and  n  different  integral  values.  The  paths  are  called  Lissajou's  curves, 
and  are  the  result  of  the  composition  of  two  harmonic  motions  with  differ- 
ent periods.  Show  that  the  whole  motion  takes  jilace  within  a  rectangle  of 
sides  2  a  and  2  h.  Are  the  curves  all  closed  or  not?  Draw  a  number  of  them 
for  m,  n  =  1,  2,  3,  •  •  • . 

31.  Show  that  in  the  motion  x  =  nfif)  +  ci-\-  d,  y  =  f'f(t)  +  mt  +  n,  the 
direction  of  the  force  is  constant  and  its  magnitude  is  \cfi  +  h^ f"^ (t) . 
Show,  conversely,  that  if  the  direction  of  a  foi'ce  is  constant,  the  trajectory 
must  have  equations  of  the  above  type. 

Hint,  x"  and  y"  are  functions  of  /.  As  their  ratio  is  constant  they  may 
both  be  regarded  as  constant  multiples  of  the  same  function. 

32.  The  enemy's  vessel  is  sailing  directly  toward  us.  Our  biggest  gun 
can  give  to  its  shell  a  fixed  initial  speed  v^^.  Considering  the  damage  done 
proportional  to  the  speed  of  the  projectile  as  it  strikes  the  enemy's  vessel, 
show  that  we  get  the  same  effect  if  we  fire  at  any  time  after  it  comes 
within  range.  Consider  in  general  terms  the  effect  of  the  air  resistance 
and  the  angle  at  which  the  shot  hits. 

33.  What  is  the  length  of  a  pendulum  beating  seconds  at  a  place  where 
^  =  32.2? 


ANALYSIS  •  77 

ANALYSIS  OF  CHAPTER  HI 

1.  Subject-matter  of  dynamics. 

2.  Rectilinear  motion.    Definitions  of  velocity  and  acceleration. 

3.  The  '<  complete  description  "  of  rectilinear  motion. 

4.  Various  special  manners  of  determining  rectilinear  motion  by  rela- 
tions involving  velocity  and  acceleration  and  the  derivation  from  them  of 
the  "complete  description."  The  student  should  be  able  to  indicate  the 
method  of  this  derivation  in  any  given  case. 

5.  Graphic  representations  of  rectilinear  motions. 

6.  Sjiecial  rectilinear  motions.  The  student  should  be  able  to  treat 
fully  the  more  important  of  tliese  motions. 

7.  The  "  complete  desci-iption  "  of  jilane  motion. 

8.  Definition  of  velocity  and  acceleration  in  plane  motion. 

9.  Some  geometric  properties  of  jilane  motion.  Tlie  student  should  be 
able  to  derive  these,  with  the  possible  exception  of  the  derivation  of  the 
values  of  the  tangential  and  normal  accelerations.  The  results  in  these 
cases  he  should  certainly  know. 

10.  The  general  differential  equations  of  plane  motion, 
ir.  Special  plane  motions.    Tlie  student  should  be  able  to  treat  fully 
the  more  important  of  these. 

12.  Definitions  of  angular  velocity  and  angular  acceleration. 

13.  The  period  of  a  simple  pendulum  and  the  determination  of  g. 

14.  Planetary  motion.    The  student  should  know  the  law  of  force,  the 
meaning  of  the  integral  of  areas,  and  the  nature  of  the  path. 


CHAPTER  IV 

WORK  AND  ENERGY 

22.  Work.    If  we  move  a  body  against  a  force,  as,  for  instance,  in 

lifting  it  against  gravity,  we  do  work  upon  the  body.    We  consider 

first  the  work  done  in  moving  a  body  in  a  straight  line  against  a 

constant  force,  for  instance  up  an  inclined  plane.    The  amount  of 

work  done  against  the  force  is  defined  to  be  minus  the  product  of 

the  distance  moved  through  hy  the  projection  of  the  force  iLpon  the 

line  of  motion,  or 

W=  —  sf,  =  —  s/cos  (s,  /). 

For  instance,  if  the  inclination  of  an  inclined  plane  be  i,  f  ==  mg 
and  cos  (s,/)  =  cos  [(7r/2)  +  i]  =  —  sint,  so  that  if  the  mass  be  moved 
a  distance  of  2  ft.,  the  work  done  against  gravity  is,  +  2mg  sin  i. 
Tlie  unit  of  work  in  general  use  is  the  engineering  unit,  the  foot 
pound,  which  is  the  amount  of  work  done  in  lifting  one  pound 
one  foot  against  gravity.  The  power  of  a  machine  is  the  time  rate 
at  which  it  can  do  work.  The  unit  of  power  is  tlie  horse  power,  or 
33,000  ft.  lb.  per  min. 

XI.    PROBLEMS  ON  WORK  AND  POWER 

1.  Find  the  horse  power  necessary  to  haul  a  train  of  200  T.  up  a  2% 
grade  at  the  rate  of  20  mi.  per  hr.,  frictional  resistance  being  2  lb.  per  T. 

Solution.  The  force  of  gravity  on  the  train  is  400,000  lb.,  and  the  com- 
ponent of  this  against  the  motion  is  400,000  sin?  where  tan  i  =  .02.  By 
tables  we  find  sint  =  .0200,  so  that  we  have  a  retarding  force  of  8000  lb. 
due  to  gravity,  and  of  400  lb.  due  to  resistance,  in  all  8400  lb.  This  is 
hauled  at  20  mi.  per  hr.,  that  is,  at  1760  ft.  per  min.  But  each  foot  means 
8400  ft.  lb.  of  work.  Thus  we  have  8400  x  1760  =  14,784,000  ft.  lb.  per 
min.,  or  448  horse  power,  about. 

2.  Calculate  the  work  done  in  raising  a  weight  of  2  T.  from  tlie  ground 
to  a  point  a  yard  above  the  ground.  Suppose  inclined  planes  of  inclinations 
45°,  80"^,  10°,  lie  used,  what  will  be  the  work  done  in  each  case,  the  final 
height  above  the  ground  being  always  1  yd.  ? 


FOECE  FIELDS  79 

3.  A  pump  discharges  water  at  the  rate  of  20  cu.  ft.  per  min.  from  an 
artesian  well  1000  ft.  deep.    What  is  its  power? 

Note.  1  cu.  ft.  of  water  weighs  62|-  lb. 

4.  A  body  of  weight  m  lb.  is  to  be  raised  from  the  ground  to  a 
heiglit  of  h  ft.  by  use  of  an  inclined  plane  making  an  angle  i  with  the 
horizontal.  If  the  coefficient  of  friction  is  fx,  show  that  the  work  done  will 
be  hm  (1  +  ficoti)  ft.  lb. 

5.  One  end  of  a  spring  whose  other  end  is  clamped  is  pulled  out  a 
distance  18  in.,  the  average  force  exerted  by  the  spring  being  3  poundals. 
What  is  the  work  done? 

6.  The  stroke  of  a  piston  whose  head  has  an  area  of  100  sq.  in.  is  13  in. 
long.  The  average  pressure  is  80  lb.  per  sq.  in.  W'hat  is  the  work  done? 
If  the  fly  wheel  makes  10  revolutions  per  min.,  what  is  the  estimated  horse 
power  ? 

7.  Show  that  the  work  done  in  moving  a  body  from  the  bottom  to  the 
top  of  a  smooth  inclined  plane  is  the  same  as  that  done  in  raising  the 
body  vertically  from  the  level  of  the  bottom  to  the  top. 

23.  Force  fields ;  work  on  curved  paths.  If  a  particle  mov- 
ing in  space  is  subject  at  each  point  to  some  definite  force  which 
is  either  constant  like  gravity  or  varies  from  point  to  point  con- 
tinuously, like  the  force  of  a  magnet  on  a  piece  of  iron,  the  particle 
is  said  to  move  in  ajield  of  force.  Work  is  done  against  the  field 
when  a  body  is  moved  against  its  forces.  Moreover,  the  path  may 
be  curved  and  the  force  may  vary.  In  order  to  extend  our  defini- 
tion of  work  given  in  the  last  section,  w^e  imagine  a  polygon 
inscribed  in  the  curved  path  and  the  body  to  be  moved  along  the 
sides  of  this  polygon.  We  shall  suppose  the  force  to  retain  along 
each  side  the  value  it  had  at  the  beginning  of  that  side.  This  will 
give  us  an  approximation  to  reality  which  is  closer  and  closer  as 
the  sides  of  the  polygon  are  decreased  in  length  and  increased  in 
number.  For  each  side  the  work  will  be  —f(s^)As-,  and  the  defini- 
tion of  the  amount  of  work  done  in  moving  the  body  from  the 
point  Sj  of  the  curve  to  a  point  s^  is  accordingly 


^1. = 1;l^o(-S/'(^')^-^') = -  j^'-^''^^ 


Ex.  1.  Show  that  the  work  done  in  raising  a  particle  against  gravity 
from  a  height  h.^  to  a  heiglit  /*„  is  independent  of  the  path  along  which  the 
body  is  raised,  and  has  therefore  the  value  (h^  —  ^i)g- 


80  WORK  AND  ENERGY 

Ex.  2.  Show  that  the  work  done  against  gravity  on  a  solid  body  by  any 
motion  whatever  is  equal  to  (h^  —  h^g  where  h■^^  and  h^  are  the  heights  at 
the  beginning  and  the  end  of  the  motion,  respectively,  of  the  center  of  mass 
of  the  body. 

Hint.  Assume  the  whole  work  is  the  sum  of  the  work  done  on  the 
separate  parts. 

Note  that  the  results  of  these  exercises  are  valid  not  only  for  gravity, 
but  for  any  force  field  where  the  force  on  the  unit  mass  is  constant  in 
magnitude  and  direction. 

XI.   PROBLEMS  ON  WORK  AND  POWER  {Continxied) 

8.  Show  that  the  work  done  in  raising  a  body  from  the  lowest  to  the 
highest  point  of  a  vertical  circle  is  the  same  whether  a  semicircle  or  the 
diameter  be  used  as  path. 

Do  this  as  an  independent  verification  of  Ex.  1,  not  as  an  apjilication. 

9.  Find  the  horse  power  necessary  to  haul  a  train  at  a  speed  of  .30  mi. 
per  hr.,  the  frictional  resistarice  being  equivalent  to  the  weight  of 
10,000  lb. 

10.  Let  X  be  the  modulus  of  elasticity  of  an  elastic  string  (i.e.  the 
tension  required  to  double  its  length,  it  being  assumed  that  increase  in 
the  length  is  proportional  to  the  tension)  and  x  its  length.  Then  the  ten- 
sion is  given  by  T  =  (x—l) X/l,  being  measured  in  pounds.  Find  the  work 
done  in  stretching  the  string  from  a  length  x  =  a  to  a  length  x  =  b. 

11.  An  automobile  ascends  a  hill,  whose  inclination  is  i  =  arctan 
(1/30),  a  distance  of  200  yd.  per  min.  If  the  automobile  weighs  1  T., 
find  the  horse  power  required. 

24.  Conservative  fields ;  potential  energy.  If  work  be  done 
upon  a  body  so  as  to  lift  it  against  gravity,  the  body,  in  turn,  by 
means  of  pulleys,  may  be  made  to  do  work  in  lifting  other  bodies. 
If  the  force  field  is  such  that  exactly  as  much  work  will  be  done 
by  the  body  in  returning  by  any  path  to  its  original  position  as 
was  done  against  it  in  moving  it  from  this  to  its  final  position,  the 
field  is  called  conservative.  Otherwise  it  is  called  nonconservative  ; 
an  example  of  a  nonconservative  force  is  friction.  Conservative 
fields  have  the  property,  which  the  student  has  proved  in  the  case 
of  gravity,  that  the  work  done  in  moving  a  hody  from,  a  'point  Jl  to 
a  point  I^  depends  only  upon  the  positions  of  these  points  and  in 
no  sense  upon  the  path  between  them. 

If  a  body  moves  in  a  conservative  field  from  a  point  J^  to  a  point 
^  against  the  forces  of  the  field,  it  acquires  a  certain  ability  to  do 


POTENTIAL  EXERGY  81 

work.  This  ability  to  do  work  is  called  energy,  and  as  it  depends 
upon  the  position  of  the  point  it  is  called  potential  energy,  that 
is,  possible  energy,  ready  to  manifest  itself  as  soon  as  the  body  is 
allowed  to  move.  It  will  readily  be  seen  that  this  energ}-  is  so 
far  a  relative  thing.  The  excess  of  energy  at  ^  over  that  at  P^  is 
defined  as  equal  to  the  amount  of  work  that  can  be  done  by  the 
body  in  passing  from  ^  to  i^,  or 


W  = 

"01 


=  -  Cfsds. 


It  is,  however,  customary  to  speak  of  the  energy  at  a  point  i^,  by 
which  we  mean  this  same  difference  JV^^,  the  point  i^  being  con- 
sidered the  standard  point  of  zero  energy.  There  will,  in  general, 
be  points  of  negative  potential  energy,  they  being  of  course  points 
such  that  in  moving  the  body  from  them  to  J^,  work  must  be  done 
against  the  field.  To  change  the  standard  point  merely  means  to 
add  a  constant  to  the  energy,  and  this  in  no  wise  inconveniences 
us  in  our  use  of  the  idea.    We  therefore  write 


w{s)=-r 


fsds 


as  the  expression  for  the  potential  energ5\* 

Ex.  What  are  the  equipotential  surfaces  of  a  field  of  central  foi'ces? 
(See  p.  68.) 

TJie  projections  on  the  axes  of  the  force  at  any  point  of  a  conserv- 
ative force  field  are  the  negatives  of  the  derivatives  of  the  potential 
energy  with  respect  to  the  corresponding  coordinates,  that  is, 

dW  ^  dW 

^'=-^'       ^'  =  ~^' 

For,  taking  two  points  on  a  parallel  with  the  a;-axis,  with  abscissas 
x^  and  x^-\- ^K^,  we  have 

*  If  all  along  the  path  used/*  =  0,  that  is,  if  the  path  is  at  all  points  i)erpendicular 
to  the  direction  of  the  force,  we  see  that  no  work  is  done  over  the  path.  All  such  paths 
radiating  from  a  point  form,  in  the  case  of  a  conservative  field,  a  surface  whose  nor- 
mals are  the  force  vectors.  Such  a  surface  is  called  an  equipotential  surface,  for  the 
potential  energy  is  the  same  at  all  of  its  points.  Thus  if  the  force  of  gravity  be  con- 
sidered constant,  the  equipotential  surfaces  are  level  planes.  From  this  fact  equipo- 
tential surfaces  for  other  force  fields  are  frequently  called  level  surfaces. 


82  WORK  AND  ENERGY 


xo  'J  ^a  %J  xi 

or,  applying  the  law  of  the  mean  for  integrals, 

Jioxj  +  Ax, 
dx  =  -  X,  {x^  +  ^A^i)  A«^, 

where  0  <  ^  <  1.    Dividing  by  Ax^  and  taking  the  limit, 

^  =  -A,     and  similarly     ^'  =  -/. 

From  this  we  derive  an  interesting  corollary  :  A  particle  is  in  equi- 
librium when  its  potential  energy  is  at  a  niaximum  or  minimum. 

The  converse,  that  if  the  body  is  in  equilibrium,  the  potential  energy  is 
at  a  maximum  or  minimum,  is  iwt  always  true,  though  frequently  so  stated. 
For  example,  if  W  —  t^  —  y'^,  f^  =fy  =  0  at  the  origin,  but  the  surface  z  = 
a;2  _  yi  ig  saddle  shaped  here,  and  has  neither  a  maximum  nor  a  minimum. 


XII.  PROBLEMS  ON  POTENTIAL  ENERGY 

1.  The  potential  energy  of  a  solid  body  or  a  system  is  a  maximum  or 
minimum  when  the  height  of  the  center  of  mass  is  a  maximum  or  minimum. 
Two  rods  of  length  I  hinged  together  like  an  inverted  V  lie  over  a  cylinder 
of  radius  a.  Let  2  d  be  the  angle  between  them  in  a  position  of  equilibrium. 
Show  that  6  satisfies  the  equation  2  a  cos  0  —  I  sin^  0. 

2.  A  right  cylinder  with  elliptic  cross  section  is  so  weighted  that  its 
center  of  mass  is  in  a  plane  halfway  between  its  parallel  bases  and  one 
quarter  of  the  way  along  its  maximum  diameter.  Find  the  possible  posi- 
tions of  equilibrium  of  the  cylinder  lying  horizontally  on  a  smooth  plane, 
and  show  that  there  are  four  or  two  according  or  not  as  the  eccentricity  is 
greater  than  l/v2. 

Hint.  This  requires  finding  the  distance  of  the  point  («/2,  0)  from  the 
point  (x,  y)  on  the  ellipse  x  =  a  cos  0,  y  =  h  sin  0  where  b"^  —  a^(\  —  e^). 

25.  Kinetic  energy.  A  body  may  have  energy  because  of 
motion.  Thus  if  a  body  is  thrown  upward  against  gravity  it  con- 
tinually gains  potential  energy  until  it  reaches  its  highest  point. 
The  energy  which  its  motion  gives  it  is  called  its  kinetic  energy, 
and  is  measured  by  the  amount  of  work  the  body  does  against  the 
forces  of  the  field  before  coming  to  rest. 

For  this  we  find  (confining  ourselves  to  the  plane) 


KINETIC  ENERGY  83 

where  s^  determines  the  position  of  rest.    But  by  IV  of  §  5  this  is 

-  /    [A  COS  {x,  s)  +  /^  COS  {y,  s)]  ds, 

,      .      dx  ,      .      dii 

or,  since  cos  {x,  s)  =  —->         cos  (y,  s)  =  -^  > 

ds  ds 


k  = 


-£yj.+fM=-£{i/-i^f.t)M-, 


and  since  7^=1^  — -  >         f.  =  m  — ^ 

•^"^  ^(5  -^^  ^^ 


or,  since  the  position  characterized  by  i^  is  one  of  rest,  v^^  =  0,  and 

we  have  ,      ,      „ 

k  =  l  mv- 

26.  Conservation  of  energy.  Suppose  a  body  move  in  a  force 
field  from  a  point  i^  to  a  point  i^;  the  gain  in  kinetic  energy  is 
\mv^—  \mv'l.  But  tliis  is  equal  to  the  work  done  by  the  forces 
of  the  field,  or  ^..^ 

/    fsds  =  -W{s,)  +  W{s,), 

so  that,  equating  and  transposing, 

that  is,  the  sum  of  the  hinetic  and  2^otential  energies  of  a  hody  is 
always  the  same  during  motion  in  a  conservative  force  field.  This 
result  is  known  as  the  principle  of  the  conservation  of  energy. 
As  the  object  of  a  perpetual-motion  machine  is  to  do  work  (at  least 
against  the  friction  of  its  own  parts)  perpetually,  always  returning 
to  some  standard  position  with  the  same  kinetic  energy,  we  see 
that  a  perpetual-motion  machine  is  hnpossihle,  provided  the  forces 
on  which  it  depends  are  conservative. 


84  WOEK  AND  ENEEGY 

Ex.  Suppose  a  body  moves  on  a  curved  path  under  the  influence  of 
gravity,  like  a  bead  upon  a  smooth  wire.  By  the  principle  that  the  energy 
is  constant,  show  that  the  speed  of  the  body  is  the  same  at  all  points  where 
it  has  the  same  heights  (cf.  problem  10  of  §  21). 

Xni.    PROBLEMS  ON  ENERGY 

[See  note  under  Suggestions  and  Answers.] 

1.  A  hammer  weighing  1  lb.  strikes  a  nail  with  a  velocity  8  ft.  per  sec, 
driving  it  a  quarter  of  an  inch  into  a  plank.  Find  the  average  force  exerted 
upon  the  nail  during  its  motion. 

Solution.  The  kinetic  energy  of  the  hammer  is  transferred  to  the  nail, 
except  a  small  amount  which  is  dissipated  in  heat,  which  we  neglect  here. 
Thus  ^m?;2  =  1.(1/30,2)82  =  .99  ft.  lb.  work  is  done  against  resistance  by 
the  nail.  If  the  force  is  constant,  we  have  for  the  work  /•  s  =/■  ^  •  Jj,  and 
as  this  is  equal  to  the  kinetic  energy,  we  have/=  48  ■  .99  =  48.3  lb. 

2.  How  much  work  is  accumulated  in  a  body  weighing  300  lb.  and 
moving  64  ft.  per  sec? 

3.  A  fly  wheel  12  ft.  in  diameter,  whose  rim  weighs  12  T.,  makes  50 
revolutions  per  min.    What  is  its  kinetic  energy? 

4.  An  automobile  running  30  mi.  per  hr.  comes  to  the  foot  of  a  hill. 
To  what  height  will  it  ascend  without  power,  friction  and  other  resistance 
being  neglected  ? 

5.  Compare  the  kinetic  energy  of  a  mass  of  20  lb.  falling  from  rest  at 
the  end  of  the  fifth  second  with  the  energy  at  the  end  of  the  sixth  second. 

6.  A  mason's  helper  throws  bricks  up  to  him  through  a  vertical  distance 
of  14  ft.,  so  that  when  the  mason  catches  them  they  have  a  vertical  velocity 
of  6  ft.  per  sec.  By  what  proportion  would  the  helper  reduce  his  work  by 
throwing  the  brick  so  as  to  reach  the  mason  with  no  speed  remaining  ? 

7.  If  a  pendulum  hanging  at  rest  is  given  an  initial  velocity  i-^,  how 
high  will  it  rise  ? 

8.  A  gun  carriage  of  2  T.  recoils  horizontally  with  a  velocity  of  12  ft. 
per  sec.  Find  the  constant  force  which  will  take  up  the  recoil  within  a 
distance  of  2  ft. 

9.  The  oOO-lb.  hammer  of  a  pile  driver  falls  10  ft.  on  to  the  head  of  a 
pile,  which  is  forced  thereby  an  inch  into  the  groimd.  Find  the  average 
force  exerted  by  the  hammer. 

10.  A  river  has  a  cross  section  of  area  16,000  sq.  ft.  and  flows  with  a 
mean  speed  of  4  mi.  per  hr.  Find  the  horse  power  that  would  be  developed 
if  all  the  energy  of  the  river  could  be  utilized. 

Note.  The  weight  of  1  cu.  ft.  of  water  is  62^  lb. 

11.  A  particle  slides  from  the  highest  point  on  the  outer  surface  of  a 
smooth  sphere  of  radius  a.  Find  the  point  where  the  particle  leaves  the 
sphere. 


a:n"alysis  85 

ANALYSIS  OF  CHAPTER  IV 

1.  Definitions  and  units  of  work  and  power. 

2.  Definition  of  work  for  a  curved  path  and  varying  force. 

3.  Definitions  of  conservative  fields  and  potential  energy. 

4.  The  force  components  given  by  the  negative  of  the  derivatives  of  the 
potential  energy  ;  equilibrium  in  terms  of  potential  energy. 

5.  Definition  of  and  expression  for  kinetic  energy. 

6.  The  principle  of  the  conservation  of  energy. 


CHAPTER   V 

MECHANICS  OF  RIGID  BODIES 

27.  Instantaneous  motion  of  a  rigid  body.  When  a  body  moves 
from  one  position  to  another,  and  we  consider  only  the  positions, 
and  leave  out  of  account  the  way  in  which  the  body  moved  from 
one  to  the  other,  we  speak  of  the  displacement  of  the  body.  The 
displacement  of  any  point  of  the  body  may  be  represented  by  a 
vector  joining  its  initial  and  final  positions.  We  next  consider  two 
special  Tnotions  of  a  body,  in  which  the  intermediate  positions  are 
considered :  namely,  translation,  in  w^hich  all  points  move  in  con- 
gruent curves,  for  which  we  may  usually  take  straight  lines,  and 
all  lines  in  the  body  remain  parallel  to  themselves ;  and  rotation, 
in  which  there  is  a  line,  or  axis,  in  the  body,  all  of  wdiose  points 
are  fixed  during  the  motion.  The  following  theorem  is  impor- 
tant: 

Any  displacement  of  a  rigid  hody  may  he  hro^ight  about  by  a 
translation  and  a  rotation.  To  see  this  we  remark  that  tlie  motion 
of  a  body  may  be  studied  by  fixing  a  set  of  coordinate  axes  in  the 
body  and  considering  their  various  positions  relative  to  a  set  of 
axes  fixed  in  space.  Let  us  denote  by  O^X^,  O^F^  O^Z^  the  initial 
position  of  the  axes  fixed  in  the  body,  and  by  O^X^,  O^Y^,  O^Z^  their 
final  position.  The  translation  desired  is,  then,  the  one  which  carries 
0^  over  into  0,.  Let  us  suppose  it  carries  O^X^,  O^Y^,  O^Z^  into 
the  parallel  set  0^X[,  0.^  Y[,  0^Z[.  .  We  have  now  to  show  that  a 
rotation  may  be  found  which  carries  0„X[,  O^Y^,  O^Zl  into  O^X^, 
0„Y„,  O.^Z^  respectively.  Let  A^,  Y^,  Z[,  X,,  Y^,  Z„,  denote  the 
points  wdiere  the  corresponding  axes  pierce  a  unit  sphere  about  the 
point  Og  (see  Fig.  31).  Then  we  need  merely  show  that  we  can  find 
a  diameter  such  that,  if  the  sphere  be  rotated  about  it  through  the 
proper  angle,  X[  will  pass  into  X^^  and  Y[  into  Y^,  for  then  Z[  wUl 

86 


DISPLACEMENTS 


87 


also  become  Z.^  because  each  Z-point  is  a  quadrant's  distance  from 
the  coiTesponding  X-  and  I'-points  and  on  the  same  side  of  the 
A'F-plane  in  both  cases.  Xow  in  this  rotation  which  we  seek,  all 
points  of  the  sphere  move  on  parallel  circles  with  the  required 
diameter  as  axis.  Hence  they  remain  unchanged  in  their  distance 
from  this  pole ;  the  pole  is  equidistant  from  X[  and  X.,,  and  is 
therefore  in  the  plane  which  perpendicularly  bisects  the  line  join- 


^z, 


Fig.  30 


ing  them.  Similarly,  it  lies  in  the  plane  which  perpendicularly 
bisects  Y[  Y^.  The  axis  is  thus  determined,  unless  the  planes  coin- 
cide. It  is  left  to  the  student  to  consider  the  meaning  of  this  case, 
and  further  to  prove  that  an  angle  can  always  be  foimd  which  will 
serve  to  bring  hotli  X[  into  X^  and  Y[  into  I';,  at  the  same  time. 

It  is  evident  that  this  resolution  of  a  displacement  into  transla- 
tion and  rotation  can  be  effected  in  many  ways,  for  instead  of 


88 


MECHANICS  OF  RIGID  BODIES 


choosing  the  particular  point  0^  as  origin  of  our  axis  in  the  body 
we  might  have  chosen  any  other  point.  It  is  important  to  notice 
that  the  translation  and  rotation  are  by  no  means  the  actual  motion 
of  the  body.  They  simply  give  a  way  in  which  it  may  be  brought 
from  its  initial  to  its  final  position.  The  real  value  of  this  resolu- 
tion consists  in  its  application  as  follows :  In  a  short  interval  of 
time,  At,  the  body  suffers  a  small  displacement,  which  may  be 
resolved  as  indicated.  The  displacement  of  any  given  one  of  its 
points  is  thus  also  resolved  into  one  due  to  the  translation  and  one 
due  to  the  rotation.  The  smaller  A^  the  more  accurately  do  these 
two  displacements  actually  represent  the  real  motion  of  the  point. 
Considering  the  displacements  as  vectors,  and  dividing  by  A^  and 

taking  the  limit,  we  see  that 
at  any  instant  the  velocity  of 
any  given  point  may  be  con- 
sidered as  decomposed  into  a 
velocity  of  translation  and  a 
velocity  of  rotation ;  or,  as  we 
frequently  express  it,  the  in- 
stantaneous motion  consists 
in  an  instantaneous  transla- 
tion and  an  instantaneous  ro- 
tation. The  important  thing, 
however,  is  that  the  velocity 
of  translation  and  the  angular 
velocity  of  the  rotation  are 
the  same  for  all  points  of  the  rigid  body.  Thus  instead  of  having 
an  infinite  number  of  velocities  to  deal  with  corresponding  to  the 
infinitely  many  points,  as  we  should  have  in  a  nonrigid  body,  we 
have  merely  two.  Similar  results  hold  for  the  acceleration.  The 
actual  finite  motion  of  the  body  is  generated  by  a  translation  with 
a  velocity  that,  in  general,  is  continually  changing,  and  by  a  rota- 
tion with  varying  velocity  about  a  varying  axis. 

The  foregoing  remarks  make  clear  the  importance  of  studying  two 
particular  cases  of  motion  of  a  rigid  body,  namely  pure  translation 


Fig.  31 


TRANSLATION  89 

and  pure  rotation.  Before  doing  so,  however,  two  general  remarks 
should  be  made. 

The  first  concerns  the  internal  forces  and  reactions  of  the  body. 
We  shall  make  the  assumption  embodied  in  Newton's  third  law  of 
motion,  that  the  forces  exerted  by  two  particles  upon  each  other  lie 
in  the  line  joining  them,  and  are  equal  in  magnitude  and  opposite 
in  direction.  This  assumption  will  enable  us  to  leave  out  of  con- 
sideration entirely  the  effect  of  internal  forces  in  the  study  of 
rigid  bodies,  as  we  shall  see  later  on. 

The  second  concerns  the  energy  of  a  system  of  a  number  of  par- 
ticles. If  two  weights  are  raised  against  gravity,  the  amount  of 
work  to  be  obtained  by  letting  them  fall  is  the  sum  of  the  amounts 
to  be  obtained  from  each  singly.  Similarly,  for  any  number  of 
particles.  We  are  thus  led  to  the  more  general  statement :  Tlie 
potential  and  Txinetic  energies  of  any  system  of  masses  are  the  sums 
of  the  potential  and  kinetic  energies,  respectively,  of  the.  individAial 
masses,  although  we  shall  not  take  the  space  to  consider  the  com- 
plete establishment  of  this  principle. 

It  would  be  a  mistake,  however,  to  assume  that  if  a  velocity  is  resolved 
into  two  component  velocities,  the  kinetic  energy  is  the  sum  of  the  energies 
due  to  the  comiwnent  velocities,  as  is  shown  by  the  simple  case  in  which 
we  consider  rest  as  the  resultant  of  two  equal  but  oppositely  directed  veloc- 
ities. If  the  sjieed  is  v,  we  should  have  for  the  sum  of  the  kinetic  ener- 
gies ^  mv'^  +  ^  7711-2,  -whereas  the  total  energy  is  obviously  0.  "NVe  shall  see 
later  w  hat  statement  can  be  made  in  this  respect  concerning  a  rigid  body 
(§  31). 

28.  Pure  translation.  The  tcork  done  upon  a  rigid  body  during 
a  translation  is  the  same  as  if  the  whole  mass  were  concentrated  at 
the  center  of  mass.  To  see  this  we  imagine  the  body  split  up,  after 
the  fashion  of  the  Integral  Calculus,  into  a  set  of  masses  which  are 
approximately  particles.  Then  all  the  elements  of  mass  Am  move 
along  congruent  curves  with  the  same  acceleration,  and  the  same 
is  true  of  tlie  center  of  mass.  If  a^  is  the  projection  of  this  com- 
mon acceleration  upon  the  common  direction  of  the  paths,  the 
projection  of  the  force  acting  upon  the  whole  mass  M  as  if  con- 
centrated at  the  center  of  mass  upon  the  direction  of  its  path  is 


90  MECHANICS  OF  EIGID  BODIES 

a^M,  and  hence  the  work  done  on  this  hypothesis  is 

—  I    a^Mds. 

As  a,  is  also  the  acceleration  of  the  mass  Awi,  the  total  force  act- 
ing upon  it  is  a^Am,  though  it  must  not  be  inferred  that  this  force 
comes  alone  from  the  external  field,  for  the  internal  stresses  of  the 
body  will  contribute  a  part.  These  forces  contribute  no  work,  how- 
ever, because  they  occur  always  in  pairs,  and  the  distances  between 
tlieir  points  of  application  remain  unchanged.    Hence  the  work  for 

the  mass  A?n.  is  -..^ 

W^  =-        a^Am,ds ; 

and  summing  and  taking  the  limit,  we  have  for  the  total  work 
W=  Hm  ^  ( —  I    a^Am^ds 

=  —  I     a^limV  AJ/,W5  =—  /    a^Mds, 

which  agrees  with  the  above  result.    In  replacing  the  summations 

by  the  integrals  which  are  their  limits,  we  eliminate  any  error 

arising  from  the  fact  that  the  masses  Am^  are  not  particles,  and 

thus  the  theorem  is  estabhshed. 

Ex.  Prove  similarly  that  the  kinetic  energy  also  of  a  rigid  hody  whose 
motion  is  a  translation  is  the  same  as  if  the  mass  were  concentrated  at  the 
center  of  mass. 

29.  Pure  rotation.  "We  ask  first  for  the  work  done  by  a  rota- 
tion, in  which  we  consider  one  force  F  as  acting,  applied  at  a  point 
(x,  y,  z).  Let  us  take  the  2-axis  along  the  axis  of  rotation.  The 
work  done  against  the  force  is,  by  definition. 


jr.= 


Let  r  denote  the  length  of  the  perpendicular  upon  tlie  axis  from 
{x,  y,  z),  (r  =y/  j?-\-  If),  and  let  6  be  the  angle  this  line  makes  with 
the  a^2^plane.  Then  if  s  is  measured  from  the  point  where  the 
circular  path  of  (./;,  y,  z)  pierces  the  a:<;-plane,  s  =  rd  and  ds  =  rdO ; 


KOTATIOX  91 

then,  as  cos {x,  s)  =  cos  \6  +  (7r/2)]  =  —  sin ^  =  —y/r,  cos {y,  s)  =  cos6 
=  x/r,  and  cos  (z,  s)  =0,/,=  /^  cos  (x,  s)  +  f^  cos  (y,  s)  +/^  cos  (z,  s)  = 
(^fv—yfJ/^>  and  we  have 

W=-   f\xfy-yf:,)de. 


Jdi 


The  quantity  xf^—  yf^  is  the  moment  of  the  force  about  the  2:-axis 
(see  §  1 1,  II,  also  Ex.  2).  Thus  the  work  done  against  a  force  hy  a 
rotation  is  the  negative  of  the  integral  of  the  moment  of  the  force  over 
the  angle  rotated  through.  If  \ve  have  a  number  of  forces  applied  at 
different  points,  g 

Here  again  the  interior  forces  may  be  left  out  of  consideration,  for 
their  moments  are  equal  and  opposite  in  pairs. 

The  kinetic  energy  of  the  mass  m  rotating  about  the  z-axis  is 
\  mv^.  But  as  s  =  rO,  v  =  ds/dt  =  r(dO/dt)  =  rco,  this  expression 
becomes  ^Tii'Tar,  and  for  a  number  of  masses 

If  our  masses  form  a  rigid  continuous  body,  we  find,  by  the  process 
of  the  Integi-al  Calculus, 

JT  =  lim  ^  ^  S  (a?..,  y^,  2;.)  rfAx,Ai/iAz,  ap- 

=  7)     / 1  f  ^  (•»'  y>  ^)  r'^dxdydz  (o\ 

Tlie  quantities 

mr^,      '^mivf,     and      /  /  |  h{x,y,  z)  rdxdydz, 

which  multiply  ^  o)^  in  the  above  expressions  for  the  kinetic 
energy,  are  called  the  moments  of  inertia  of  the  particle,  system 
of  particles,  or  the  continuous  body  respectively.  It  will  be  noticed 
that  if  we  consider  angular  velocity  in  rotation  as  corresponding 
to  speed  in  translation,  they  occupy  exactly  the  same  place  with 
respect  to  the  energy  of  rotation  that  the  mass  does  with  respect 
to  the  energy  of  translation ;  they  are  the  coefficients  of  half  the 
angular  speed ;  they  measure  the  inertia  or  resistance  of  the  body 


92 


MECHANICS  OF  KIGID  BODIES 


to  turning  forces.  This  fact  makes  them  of  gi-eat  importance  both 
in  the  theory  of  rotating  bodies  and  also  in  other  problems  of 
engineering,  where  similar  expressions  find  application  to  the 
theory  of  strength  of  materials.  We  shall  therefore  devote  a 
separate  paragraph  to  them. 

30.  Moments  and  products  of  inertia.    Radius  of  gyration.    In 
the  integral 

/  =  III  8  (x,  y,  z)  r'^dxdydz 

the  limits  are  determined  just  as  in  the  volume  problems  of  the 
Integral  Calculus,  by  the  boundaries  of  the  body  considered.  The 
expression  holds  for  any  axis,  where  r  denotes  the  distance  of 
the  point  x,  y,  z  from  that  axis.    In  particular 

A=  fCCs  {x,  y,  z)  (y^  +  z')  dxdydz, 
B  =   fffs  (x,  y,  z)  (2-2  +  x2^  dxdydz, 

C  =  fff^(x,  y>  ^) (^'  +  y') dxdydz, 

are  the  moments  of  inertia 
about  the  three  coordinate 
axes.  If  we  have  any  other 
axis  a,  the  distance  r  of  the 
point  {x,  y,  z)  from  this  axis 
is  one  side  of  a  right  tri- 
angle, of  wliich  the  hypot- 
enuse OP  is  the  distance 
Vx"^  +  y'^  +  ^  of  {x,  y,  z) 
from  the  origin,  and  one 
side  ON  is  the  projection* 

X  cos  {x,  a)  +  y  cos  {y,  a)-\-z  cos  {z,  a)  of  this  distance  upon  the  axis 

a.    We  have,  therefore, 


Fig.  32 


*  This  is  merely  an  extension  to  space  of  IV  of  §  5.    OP  is  a  vector  with  projec- 
tions x,y,zon  the  axes. 


MOMENTS  OF  INERTIA  93 

=  \x-  +  if-  +  r]  —  \_x  cos  (x,  a)+  1/  cos  (i/,  a)  +  z  cos  (z,  a)Y 
=  a;^[l— cos^(a.',a)]+  y"[l—  cos^(y,  a)]  +  z^[l—  cos^(z,  a)] 

—  2yz  cos (?/, a) cos (2, a)—2zx  cos (2;, a) cos {x, a) 

—  2  xy  cos  (^,  a)  cos  (2/,  a) ; 

or,  remembering  that  cos^(x',  a)  +  cos^(y,  a)+  cos^(2,  a)  =  1, 

r^  =  cos^  {x,  a)  [f  +  2^]  +  cos'^  (y,  a)  [2^  +  x^]  +  cos^  (2,  a)  [x^  +  /] 

—  2  cos (y,  a) cos (z,  a)— 2  cos (2,  a) cos (x,  a) zx 

—  2  cos  (x,  a)  cos  {y,  a)  yx. 

Multiplying  by  h{x,  y,  z)  and  integrating,  and  remembering  that 
cos(a:;,  a),  cos(y,  a),  cos  (2,  a),  being  the  cosines  of  the  angles  be- 
tween the  axis  a  and  the  coordinate  axes  are  therefore  constant, 
we  find 


=///^<^'^' 


z)  r^dxdydz 


=  A  cos2 (x,  a)  +  B cos^ (y,  a)  -\- C cos^ (z,  a)  —  2D cos {y,  a) cos (z,  a) 
—  IE cos {z,  a) cos {x,  a)  —  2F cos {x,  a) cos (y,  a),  (1) 

where  D=  III  S(x,  y,  z)yzdxdydz, 

E  =  III  S{x,  y,  z)  zxdxdydz, 
F  =  III  S(x,  y,z)xy  dxdydz, 

are  the  so-called  products  of  inertia.  Thus  having  calculated  the  six 
coefficients  of  inertia,  A,  B,  C,  and  D,  E,  F,  we  can  find  from  them 
the  moment  of  inertia  about  any  axis  through  the  origin  by  means 
of  the  formula  (1)  above.  The  coefficients  are  usually  calculated 
for  coordinate  axes  with  origin  at  the  center  of  mass.  The  follow- 
ing theorem  then  enables  us  to  find  the  moment  of  inertia  about 
any  axis,  even  if  not  through  the  center  of  mass,  without  further 
integrations,  unless  we  need  one  to  find  the  mass:  The  moment  of 
hiertia  about  any  yiven  axis  is  equal  to  the  mo^nent  about  a  parallel 


94 


MECHANICS  OF  RIGID  BODIES 


axis  through  the  center  of  mass  increased  hy  the  moment  of  inertia 
about  the  given  axis  of  the  hody  considered  concentrated  at  its  center 
of  mass. 

In  symbols,  if  /'  denote  the  required  moment  of  inertia  and  /  the 
moment  about  a  parallel  through  the  center  of  mass,  and  a  the 
distance  between  the  parallel  axes, 

/'  =  /  +  Md".  (2) 

To  prove  this,  we  take  our  coordinate  axes  through  the  center  of 

mass,  so  that  the  2;-axis  is  parallel  to  the  given  axis,  and  so  that 


Fig.  33 


the  «2;-plane  passes  through  it.  The  equations  of  the  given  axis 
will  then  be  a;  =  a,  y  =  0.  If,  then,  r'  denote  the  distance  of  the 
point  {x,  y,  z)  from  the  given  axis,  we  have 


r' 


(a-  —  a)-+  y'^=  x^-\-  'f—  2  ax  +  a^—  r^+  a^—  2  ax. 


Hence 
I 


a^  I  I  j  8  (x,  y,  z)  dxdydz  —  2  a  I  I  I  8  {x,  y,  z)xdxdydz 


+ 


The  first  integral  on  the  right  is  /,  the  second  M,  and  the  third 
is  xM  (see  §  13).  But  as  the  origin  was  taken  at  the  center  of 
mass,  ^  =  0,  so  that  this  term  drops  out  and  equation  (2)  is  thus 
established. 


PEOBLEMS  OX  MOMENTS  OF  INERTIA  95 

If  the  whole  body  were  concentrated  on  an  axis,  the  moment  of 
inertia  about  that  axis  would  vanish.  We  ask,  How  far  from  the 
axis  should  the  mass  of  the  body  he  concentrated  in  order  that  its 
moment  of  inertia  with  respect  to  that  axis  he  the  same  as  when  the 
body  has  its  given  form  ?  The  distance  sought  is  called  the  radius 
of  gyration  and  is  denoted  by  B.    From  its  definition 


MK^  =  L     Hence     R=\  — 


M 

XIV.   PROBLEMS  ON  MOMENTS  AND  PRODUCTS  OF  mERTIA  AND  RADH 

OF  GYRATION 

Unless  otherwise  specified,  take  the  density  (x,  y,  z)  equal  to  1.  Choose 
the  coordinate  axes  as  symmetrically  as  possible.  Calculate  A,  B,  C,  D, 
E,  F  and  the  radii  of  gyration  about  the  coordinate  axes.  Straight  wires 
should  be  laid  along  the  x-axis,  and  plates  in  the  xy-iplane. 

The  student  should  observe  that  sometimes  we  can  tell  in  advance  of  a 
definite  integral  that  it  vanishes,  namely,  in  cases  where  the  field  of  inte- 
gration is  symmetric  with  respect  to  a  point,  line,  or  j^lane,  and  the  inte- 
grand has  values  which  are  equal  in  absolute  value  but  opposite  in  sign  at 
pairs  of  symmetric  points.    Thus 


xdz  =  0,       I       afdx  =  0,       j    coaxdx  =  0, 

-1  J -a  Jo 


££ 


Vcl? 


xyflydx  —Q,       f     f     f  sin t|-  +  t  +  -) dzdydx  =  0. 
J -aJ-bJ-c  \a       h       cj 


The  truth  of  this  statement  is  obvious  when  we  recall  the  fact  that  the 
integral  is  the  limit  of  a  sum. 

Two  bodies  with  the  same  coefficients  of  inertia  are  sometimes  'called 
dynamlcnlly  equivalent. 

1.  Calculate  the  coefficients  of  inertia  of  an  equilateral  triangular  plate 
of  side  a,  and  of  unit  surface  density ;  also  the  moment  of  inertia  about  a 
side,  about  an  axis  perpendicular  to  the  jtlate  and  through  one  corner,  and 
finally  about  an  axis  through  one  corner  and  making  an  angle  of  45°  with 
the  plate  and  lying  in  the  plane  that  peq^endicularly  bisects  the  oii^wsite 
edge  of  the  plate.    Give  also  the  radii  of  gyration. 

Solution.  The  center  of  mass  is  at  the  intersection  of  the  medians,  which 
we  take  for  origin  of  coordinates,  taking  one  median  along  the  y-axis.  We 
find  the  equations  of  the  sides  to  be  3  x  +  v3  y  —  a  =  0,  3x  —  VS  y  +  a  =  0, 
and  2  Vs  y  +  a  =  0.    Then 

A=ff(y^  +  z^)dm, 


96  MECHANICS  OF  RIGID  BODIES 

and  since  we  have  a  plate,  2  =  0,  and  with  unit  density  dm  =  dxdy,  so  that 
we  have  ,_ 

,  JL  o  -  V  3  J/ 


A  =ffy^dxdi,  =   f^'     C      '^  y'^dxdy  =   C^'  ymn  -^iy)dy 
2V3  3  2Vi 

~3[3  4     J  _^.  "3273' 


1  /«  -  Va  w\M   ^       2     1 1 "  "*"  2 


V3    4\       3       /    I ^      V3    4\    3    /       32V3 

2  Vs 


C=  ff  (x2  +  y^)dxdy  =  A  +B  =  -^—=  ■ 


V3 

Moreover,  D  =ffyzdxdy  =  0,  and  E  =f fzxdxdy  =  0  since  z  =  0,  and 
F  =  f  fxydxdy  =  0  because  at  points  of  the  plate  that  are  symmetric  with 
resi>ect  to  the  y-axis,  (x,  y)  has  numerical  values  but  opposite  signs. 

To  find  the  moment  of  inertia  about  a  side,  we  select  the  side  that  is 
parallel  to  the  x-axis  and  employ  equation  (2).  The  x-axis  passes  through 
the  center  of  mass  and  the  moment  about  it  is  ^4.  Its  distance  from  the 
side  is  a/2  v  3.    Hence  for  a  side 

the  mass  being  the  same  as  the  area.    This  gives 

a*  V3  a^  _  V3  g* 

~32V3       16-3  ^     32 

The  same  method  gives  us  the  moment  about  a  perpendicular  through 
a  corner.    Take  the  corner  on  the  y-axis.    We  have,  then, 


\V3/  \  4  /  16  V3  4V3  16  V3 
Finally,  to  find  the  moment  about  the  axis  through  a  corner,  and 
inclined  at  an  angle  of  45°  to  the  plate,  we  find  first  by  equation  (1)  the 
moment  about  the  parallel  axis  through  the  center  of  mass.  We  have 
for  this  axis  the  angles  (x,  a)  =  90°,  (y,  o)  =  13.5°,  (2,  a)  =  45°,  and  hence 
cos(x,  a)  =  0,  cos(y,  «)  =  — I/V2,  cos(c,  o)  =  l/V2.  Whence  for  this 
axis,  by  equation  (1),  ^      ^      ^^^^4 

^=2  +  2==  ~6r' 


PROBLEMS  ON  MOMENTS  OF  INERTIA  97 

and  it  remains  to  find  the  moment  about  a  parallel  axis,  whose  distance, 
measured  along  the  plate,  is  a/ v  3,  but  measured  along  a  common  perpen- 
dicular is  a/Vs  cos  45°  =  a/V6.    Hence  for  the  required  axis  through  the 

V3«4  .  /_a_\VV3    A       11a* 


7  =  ^7^+1 


64        VVg/  \  4       /      64  V3 


The  radii  of  gyration  are  all  found  by  the  formula  R  =  Vj/M,  and  are 
respectively  :  about  OX,  a/2Vo  ;  about  OY,  g/'IVq  ;  about  OZ,  a/2V3  ; 
aboiit  a  side,  rt/SV 2  ;  about  a  perpendicular  through  a  corner,  5  a/2  v3  ; 
about  the  slanting  axis,  "vlla/4V'3. 

2.  Show  that  for  straight  wires  A  =  0,  B=C,D==E  =  F=0. 

3.  Show  that  for  plates  C  =  A+B,D  =  E=^0. 

4.  Show  that  the  moment  of  inertia  of  a  body  about  an  axis  through 
the  center  of  mass  is  less  than  about  any  parallel  axis. 

5.  Calculate  B  for  a  straight  homogeneous  wire  of  length  I.  Calculate 
its  moment  of  inertia  about  an  axis  through  its  center  of  mass  and  inclined 
at  45°  to  it.  Calculate  independently,  and  by  equation  (2)  of  the  present 
paragraph,  the  moment  of  inertia  about  a  perpendicular  axis  through 
one  end. 

6.  Calculate  the  coefficients  of  inertia  for  a  uniform  square  plate  of  side  I. 
Find  the  moments  of  inertia  about  a  diagonal  and  about  a  perpendicular  to 
the  plate  through  one  corner  ;  also  about  a  perpendicular  through  the  mid- 
point of  one  side. 

7.  Calculate  the  coefficients  for  a  homogeneous  cube  of  side  I.  Calcu- 
late the  moments  of  inertia  about  an  edge  and  about  a  diagonal.  (For  the 
diagonal  cos  (x,  a)  =  cos(,?/,  o)  =  cos  (2,  a),  and  as  cos2(x,  a)  +  cos^(y,  a)  + 
cos^(z,  a)  =  1,  each  can  easily  be  found.)  Show  that  for  all  axes  through 
the  center  of  mass  I  is  the  same. 

8.  Calculate  the  coefficients  of  inertia  of  a  homogeneous  circular  plate 
of  radius  r.  What  must  be  the  radius  of  the  plate  to  have  the  same  coeffi- 
cients as  the  square  of  problem  5  ?  The  square  is  dynamically  equivalent 
to  this  disk. 

9.  Calculate  the  coefficients  of  inertia  of  a  homogeneous  sphere  of  radius 
r.  What  must  7-  be  in  order  that  the  coefficients  agree  with  those  of  the 
cxihe  in  problem  6?  Calculate  the  moment  about  a  line  through  the  center 
with  direction  angles  (x,  a),  (ij,  a),  (z,  «).  The  cube  is  dynamically  equiv- 
alent to  this  sphere. 

10.  Calculate  the  coefficients  for  a  wire  ring  of  radius  r. 

11.  Calculate  C  for  an  arc  corresponding  to  an  angle  2  a  of  the  above 
wire  ring,  the  origin  being  at  the  center  of  mass. 

12.  Calculate  the  coefficients  of  inertia  of  the  elliptic  plate  x'^/a'^  + 
y2//;2  —  I  Calculate  its  moments  of  inertia  about  axes  through  its  focus 
parallel  with  its  minor  axis  and  perj^endicular  to  its  plane. 


98  MECHANICS  OF  EIGID  BODIES 

13.*  Calculate  the  coefficients  of  inertia  of  the  homogeneous  ellipsoid 
x2/a2  +  y^/b"^  4-  z^/c^  =  1. 

14.  Show  that  if  a  homogeneous  body  is  symmetric  with  respect  to  the 
xy-plane,  Z)  =  0.  What  further  similar  statements  can  you  make  regarding 
bodies  with  one  or  more  planes  of  symmetry  ? 

15.*  Find  the  moment  of  inertia  of  the  cardioid  plate  bounded  by 
/»  =  o  (1  —  cos  ^)  about  an  axis  through  the  origin  and  perpendicular  to  its 
plane  ;  also  about  the  axis  of  symmetry  in  the  plane. 

16.  Determine  the  coefficients  of  inertia  of  a  cuboid  of  dimensions  o,  h, 
and  c  ;  also  moments  of  inertia  about  edges  and  about  a  diagonal. 

17.  Find  the  moments  of  inertia  of  the  above  cuboid  about  the  diagonals 
of  the  rectangles  forming  its  faces. 

18.*  Find  the  moment  of  inertia  of  the  elliptic  plate  //;•  =  1  —  ecos^ 
about  a  perpendicular  through  the  focus  (which  is  the  origin  as  the  e<pia- 
tion  is  written).  Compare  the  result  with  problem  12,  noting  that  e^  = 
(a2  -  &2)/«2  and  I  =  V^/a. 

19.  Find  the  moment  of  inertia  about  its  axis  of  a  right  circular  cone 
of  altitude  a  and  radius  of  base  r. 

20.  Find  the  coefficients  of  inertia  of  a  right  circular  cylinder  of  height 
a  and  radius  of  base  r. 

21.  Calculate  the  coefficients  of  inertia  of  a  flat  angle  iron  as  indicated 
in  Fig.  17,  p.  39,  the  origin  being  the  center  of  mass. 

22.  Do  the  same  for  the  plates  of  Figs.  14  and  15,  taking  simple  numer- 
ical values  for  the  constants. 

23.  What  is  the  effect  on  the  moments  of  inertia  and  the  radius  of 
gyration  of  a  homogeneous  body  if  the  density  is  multiiDlied  by  a  constant 
k,  that  is,  increased  in  the  ratio  \:k1 

24.  In  problem  5  calculate  the  moment  of  inertia  of  the  wire  about  an 
axis  through  its  mid-point  and  making  an  angle  Q  with  it.  How  does  the 
moment  of  inertia  change  as  Q  varies  from  0  to  7r/2  ?  Can  you  explain  this 
from  mechanical  considerations? 

25.  Given  a  circular  plate  of  radius  r  from  which  have  been  removed 
four  circular  pieces  of  radius  r'  with  centers  at  the  mid-points  of  four 
equally  spaced  radii  of  the  plate.  What  is  the  value  of  r'  if  the  radius  of 
gyration  of  the  plate  about  an  axis  perpendicular  to  its  plane  through  its 
center  is  |^  the  radius  of  the  plate  ? 

26.  Suppose  a  body  expand  in  such  a  way  that  it  always  remains  similar 
to  itself  in  form  and  that  its  mass  is  constant  in  amount.  How  are  its 
coefficients  of  inertia  and  its  radii  of  gyration  affected  ? 

31.  Work  and  energy  in  the  case  of  a  rigid  body.  Let  us  con- 
sider first  a  system  of  particles  m,^  with  coordinates  (.>;,-,  ?/,•,  2,)  with 

*  The  reckonings  in  the  starred  problems  are  rather  protracted.  Integration  tables 
should  be  used,  e.g.  Peirce  (formulas  308  and  300  for  problem  18). 


WORK  OK  A  EIGID  BODY 


99 


respect  to  a  fixed  set  of  axes.  As  we  are  about  to  study  how  the 
work  and  energy  of  the  body  depends  upon  the  work  and  energy 
due  to  translation  and  rotation  separately,  we  shall  introduce  also 
a  system  of  axes  moving  parallel  with  the  old,  with  origin  always 
at  the  center  of  mass  x,  y,  z.  If  («/,  yl,  zl)  are  the  coordinates  of 
a  point  with  respect  to  these  axes,  we  have,  by  Analytic  Geometry, 


lAy-  -~~    tAj    ^\        tA-  • 

z,  =  z  +  z[  J 
The  work  done  against  the  force  field  in  moving  m,.  will  be 


(1) 


[/«  cos  {x^,  s.)  +  /^,  cos  {y„  s.)  +/,,  cos  {z„  s,)](Zs.., 


or,  as  cos  {x^,  s,)  =  — -  »  cos  (y,-,  s,)  =  -~  >  cos  (s.,  s-)  =  —^  > 

CtS-  CtS-  (-l/Oi 


X%[        dx^  dy^  dz^ 

I  Jxi  /!„       I    Jyi  A„       I    Jzi  ,7„ 


ds 


'^r      ilx.  dy.  dz, 

J  xi    ,lj.    '^  J  yi    ,li      'Jzi    Ji 


ds, 

dx. 
'dt 


y^  ds, 
dt 


or 


-XT 

^-  =  -  f\fA  +  /A  +fziK,]  dt. 


ds, 
dt, 


Now,  differentiating  (1)  with  respect  to  t,  we  have 

V  ■=^  V-+  v'  ■ 

XI  X     '  xt 


yi  1/    '  in 


and  using  these  values,  we  have 


w,=-  r[fxiV,+f,cv,+A,v,]dt-  ru.<.+^<+^<] 


(2) 


(3) 


dt. 


If  now  we  have  a  number  of  particles,  the  work  done  against  them 
all  is  the  sum  of  the  work  done  against  them  separately  (see  p.  89). 
In  summing  we  should  recall  that  v-^,  v-,  v-,  are  the  same  for  every 


100  MECHANICS  OF  RIGID  BODIES 

term  in  the  sum,  so  that  we  simply  add  their  coefficients.    The  sums 

x=^XL,,    y=XU    z=%f„ 

i  i  i 

are  the  projections  on  tlie  axes  of  the  total  forces  acting  on  the 
system.    The  result  is 

w=-  r\Ax+Yv^+zv,]dt-x  rVx.<+/x+/x]<  (4) 

in  which,  by  comparing  with  (2),  we  see  that  the  second  term  is 
the  total  work  computed  as  if  the  center  of  mass  were  at  rest,  for 
v'^,  v'^,  v^.,  are  the  velocities  relative  to  the  center  of  mass. 

The  considerations  apply  in  particular  to  rigid  bodies.  In  either 
case  the  theorem  is  true :  The  work  done  in  moving  a  rigid  lody 
is  eqtoal  to  the  work  done  in  moving  the  center  of  mass  computed 
as  if  all  the  forces  were  applied  there,  plus  the  ivhole  work  done 
on  the  hody  computed  as  if  its  center  of  mass  were  at  rest. 

In  a  conservative  force  field,  inasmuch  as  the  work  done  is  for 
each  particle  the  same,  no  matter  what  path  is  taken,  so  for  a  rigid 
body  the  work  depends  only  upon  the  initial  and  final  positions. 
As  the  change  may  be  brought  about  by  a  translation  and  a  rota- 
tion about  the  center  of  mass,  we  may  say  that  in  moving  a  rigid  hody 
against  a  conservative  force  field,  the  work  done  is  the  sum  of  the 
work  done  in  translating  its  center  of  mass,  the  forces  being  regarded 
as  concentrated  there,  plus  the  work  done  in  rotating  the  body  about 
its  center  of  mass. 

Similar  results  hold  for  the  kinetic  energy.  We  have  for  the 
particle  m,-, 

h=\^ri,vf=\m,{vl,-\-vl,+  vi;\ 

When  we  add  for  all  particles,  the  terms  in  the  bracket  lead  to 
three  sums  like 

Xm;V-v'-  =  V-  ^  m;v'-  =  v-  ^  m.  —^  =  ■?•---  "S^  m.xL 

i  i  i  I 

But  Sm,a;(.  is  nothing  else  than  M  times  the  x  of  the  center  of 


_  1 


=  km 


PROBLEMS  ON  RIGID  BODIES  101 

mass  referred  to  the  same  axes  as  x'l,  y\,  z\,  that  is,  to  axes  through 
the  center  of  mass.    It  is  therefore  0,  and  we  have 

i 

and  extending  to  rigid  bodies,  we  may  say  that  the  kinetic  energy  of 
a  rigid  body  is  the  kinetic  energy  computed  as  if  the  body  were  con- 
centrated at  its  center  of  mass,  pins  the  kinetic  energy  of  the  motion 
relative  to  the  center  of  mass  coinputed  as  if  the  point  were  fixed. 

We  have  seen  (p.  88)  that  the  instantaneous  motion  of  a  body 
may  be  resolved  into  the  instantaneous  translation  of  one  of  its 
points,  for  which  we  here  take  the  center  of  mass,  and  a  corre- 
sponding instantaneous  rotation.  If  a>  is  the  instantaneous  angular 
velocity  and  R  the  radius  of  gj^'ation  about  the  instantaneous  axis, 
we  have  for  the  kinetic  energy, 

k  =  \mv^^\  mRW, 

where,  in  general,  v,  E,  (o,  and  the  axis  of  rotation  are  continually 
varying. 

XV.   PROBLEMS  ON  WORK  AND  ENERGY  IN  THE  CASE  OF  RIGID  BODIES 

The  following  problems  are  intended  mainly  as  applications  of  the 
principle  of  the  conservation  of  energy. 

1.  Study  the  motion  of  a  homogeneous  sphere  of  mass  M  and  radius  a 
rolling  down  an  inclined  plane  of  height  h  and  inclination  i.  Show  that  at 
all  points  the  speed  of  its  center  is  less  than  if  it  slid  down,  in  the  ratio  of 
VorVz.  (Note  that  for  a  sphere  of  radius  r  rolling  so  that  its  center 
describes  a  straight  line,  v  —  rw.) 

2.  Show  that  the  time  of  rolling  down  the  incline  is  Vl4  h/5g-  cosec  i. 

3.  Show  that  for  a  homogeneous  sphere  the  times  of  rolling  down  the 
chords  of  a  vertical  circle  joining  the  highest  point  of  the  circle  to  other 
points  of  the  circumference  are  all  the  same. 

4.  A  man  has  a  hollow  iron  ball  and  a  solid  aluminum  ball  of  the 
same  radius  and  weight.  They  are  painted  so  that  they  appear  the  same. 
Explain  how  he  can  distinguish  them  by  means  of  an  inclined  plane,  telling 
which  is  which. 

5.  A  pendulum  is  formed  by  a  homogeneous  ball  rolling  on  a  circular 
track,  the  plane  of  the  circle  being  vertical.  Show  that  the  motion  is  that 
of  the  ordinary  simple  pendulum  whose  length  is  7/5  times  the  radius  of 
the  path  of  the  center  of  the  ball. 


102  MECHANICS  OF  RIGID  BODIES 

6.  Show  that  if  a  homogeneous  right  circular  cylinder  roll  down  any 
incline,  the  ratio  of  the  kinetic  energy  of  rotation  to  the  kinetic  energy  of 
translation  is  constant.  What  is  the  constant  ratio?  Show  that  a  similar 
statement  holds  for  a  homogeneous  sphere  rolling  on  any  track. 

7.  In  problem  11  of  §  26  a  particle  sliding  down  a  sphere  was  considered. 
Change  this  by  asking  at  what  point  a  rolling  sphere  of  radius  r  leaves  the 
surface  of  the  given  sphere  of  radius  a. 

8.  The  moon  rotates  about  the  earth,  keeping  the  same  side  always 
toward  the  earth.  Suppose  that  it  always  faced  the  same  direction  in 
space,  would  it  have  more  or  less  kinetic  energy  ? 

32.  Compound  pendulum  ;  experimental  determination  of  mo- 
ments of  inertia.  In  the  simple  pendulum  we  have,  theoretically, 
a  particle  moving  in  a  circular  path.  We  now  consider  any  heavy 
body  free  to  turn  under  the  influence  of  gravity  about  a  horizon- 
tal axis.  Such  a  body  is  called  a  compotind  pendulum.  If  0  is  the 
angle  between  the  vertical  plane  and  the  plane  containing  the  cen- 
ter of  gravity,  both  planes  passing  through  the  axes,  we  have  for 
the  kinetic  energy, 

,       1     /ddV     1      ,      1  ^^^,  2 
2    \dt/      2  2 

while  the  potential  energy,  or  work  done,  is  3f(/h,  h  being  the 
height  through  which  the  center  of  mass  is  raised.  Let  d  be  the 
distance  of  the  center  of  mass  from  the  axis.  Then  h  =  d  —  d  cos  6, 
and  as  the  sum  of  the  energies  is  constant, 

1  MRW  -I-  Mgd  ( 1  -  cos  ^)  =  c. 

Let  us  suppose  the  pendulum  falls  from  an  angle  6^.  At  this  point 
ft)  =  0,  so  we  have 

0  +  Mgd  {I—  cos  6^)  =  c. 
Hence,  eliminating  c, 

\  MR^'J-  =  M(jd  (cos  d  -  cos  6^, 


or 


[di)=-^^'''^-'''^^^- 


Comparing  this  with  the  formula  for  the  simple  pendulum,  (c)  of 
§21,  we  see  that  the  motions  are  the  same,  provided  2gd/R^  = 
2  g/l,  or  provided  I  =  R^/d ;    that  is,  the   compound   pendulum 


COMPOUND  PENDULUM 


103 


.Point  of  suspension 


Center  of  mass 


behaves  just  like  a  simple  pendulum  of  length  R'^/d.    For  small 
oscillations  the  period  is 

As  the  position  of  the  center  of  mass  of  a  body  can  usually  be 
simply  determined,  particularly  if  the  body  have  symmetry,  and 
hence  d  may  be  found,  and  as  T  can  be  readily  observed,  the  above 
formula  determines  the  radius  of  gyration  R,  and  hence  also  the 
moment  of  inertia  about  the  axis 
of  suspension. 

The  point  in  a  line  perpendicu- 
lar to  the  axis  of  suspension  and 
through  the  center  of  mass,  and 
distant  I  from  the  axis,  is  called 
the  center  of  oscillation,  and  is 
the  point  at  wliich  the  weight  of 
the  equivalent  simple  pendulum 
would  be  concentrated.  Let  its 
distance  from  the  center  of  mass 
bee?'.  Let  us  take  an  axis  through 
it  parallel  with  the  axis  of  sus- 
pension, and  call  the  radius  of 
gyration  about  this  axis  R'.  If  R^ 
be  the  radius  of  gyration  of  the 
body  about  a  parallel  axis  through 

the  center  of  mass,  we  have  seen  (§  30,  (2),  remembering  that 
I=MR-)  R'  =  R^+d''  and  R''^  =  Rf -{■  d'\  so  that,  subtracting, 
R'^  —  R'-^  =  d^  —  d''\  Suppose  now^the  body  be  suspended  by  the 
parallel  axis  through  the  center  of  oscillation,  and  let  Z'  be  the 
length  of  the  equivalent  simple  pendulum.  As  /  =  R'^/d,  R^  =  Id, 
and,  similarly,  R'^  =  I'd',  the  above  equation  gives 
ld-l'd'  =  {d  +  d'){d-d'). 

But  d  +  d'  =  I.    Hence  Id  -  I'd'  =l{d-  d'),  or  I'd'  =  Id',  and  since 
d'  =^Q,  1  =  1',  that  is,  the  body   suspended  from  either  axis  is 


Fig.  34 


104  MECHANICS  OF  KIGID  BODIES 

equivalent  to  the  same  simple  pendulum.  In  other  words,  point  of 
suspension  and  center  of  oscillation  may  he  interchanged,  it  being 
understood  that  the  axes  used  are  parallel. 

33.  General  equations  of  motion  of  a  rigid  body.  Any  set  of 
magnitudes  which  determine  the  position  of  a  body  are  called  the 
coordinates  of  the  body.  The  motion  of  the  body  is  determined 
when  these  coordinates  are  determined  as  functions  of  the  time.  If 
the  motion  is  given  through  the  forces  that  act,  we  shall  have  differ- 
ential equations  from  which  to  determine  these  functions,  in  general 
one  differential  equation  for  each  coordinate.  Let  us  ask  how  many 
coordinates  (and  hence  how  many  differential  equations)  are  neces- 
sary to  fix  the  position  (or  the  motion)  of  a  rigid  body.  To  fix  one 
point  demands  three  coordinates.  The  body  is  then  free  to  turn 
about  the  point.  To  fix  the  direction  of  a  line  through  the  point 
demands  two  more  magnitudes,  for  instance,  two  of  the  three  direc- 
tion cosines  (the  third  may  be  found  from  the  fact  that  the  sum  of 
their  squares  is  unity).  The  body  is  then  free  to  turn  about  an  axis, 
and  if  the  angle  through  which  it  may  be  supposed  to  have  turned 
is  fixed,  the  body  is  fixed  also.  In  all,  then,  there  are  six  coordi- 
nates, which,  however,  might  have  been  selected  in  a  multitude  of 
other  ways.  We  must  therefore  establish  six  differential  equations, 
and  we  shall  briefly  indicate  how  this  may  be  done. 

If  we  have  a  set  of  particles  m,.,  the  coordinates  of  each  satisfy 
the  equations  ,0. 


d'^y 

"^^-de^^^ 

dh       . 
'dt^      -^^ 


(1) 


where  the  quantities  f^,  /y,.,  f^^  embody  all  forces,  not  only  external 
ones  but  the  reactions  of  the  particles  on  each  other.  The  reactions 
we  endeavor  to  eliminate.  Our  first  step  is  to  write  down  the  first 
equation  (1)  for  each  body  and  add  them  all;  similarly,  for  the 
other  two  equations.     In  a  rigid  body,  thought  of  as  a  set  of 


EQUATIONS  OF  MOTION  105 

particles,  the  reactions  between  two  particles  are  equal  and  oppo- 
site in  sign,  and  hence  drop  out  in  the  sum.  So  if  X,  Y,  Z,  represent 
the  projections  on  the  axes  of  the  resultant  of  all  the  external 
forces,  we  have 

2™-rf?'=^'  2;™.J=j'  2;™.^'=^. 

or,  as  ^  m-Xf  =  Mx, 

where  31  is  the  total  mass,  and  x  the  x  of  the  center  of  mass. 


df 

dt' 
d^z 

dt'         J 


(2) 


that  is,  the  center  of  mass  moves  as  if  the  whole  mass  loere  con- 
centrated there  and  all  the  external  forces  acted  there. 

This  gives  us  three  of  our  differential  equations  of  motion. 
Another  way  to  eliminate  the  effects  of  internal  forces  is  suggested 
b)"  the  definition  of  the  moment  of  a  force  about  an  axis  as  the  prod- 
uct of  its  magnitude  by  the  perpendicular  upon  its  line  from  the 
axis  (see  §  9).  As  the  internal  forces  pair  off  into  equal  and  oppo- 
site forces  acting  along  the  same  lines,  their  total  moments  about 
any  axis  must  vanish.  Hence  we  form  the  moments,  first  of  the 
forces  acting  upon  m-  and  about  the  2;-axis.    Calling  it  N^,  we  have 

/    dSj,         d^x\      d       I    dij,         d.r, 


^  •  df      -"  dt^  j     dt    \'  dt      *"  dt 
Tlie  expression 

^'  'dt  ~  ^'  iii  ^  ^'w  ^'"  ~  f  ^"")  ^  ''•■  '^^""'  ^^^  ^^"  ^'^  "^  *"'^'  ^°^  ^^"  ^•■^-' 


where  s.  denotes  the  direction   perpendicidar  to  the  radius  r,. 
This  is  called  the  moment  of  the  velocity  about  the  z-axis,  and 


106  MECHANICS  OF  EIGID  BODIES 

the  mass  times  the  moment  of  the  velocity  is  called  the  moment 
of  momentum.    We  may  therefore  interpret  the  equation 

d 


\  '  dt    ^'  dtj 


■dt"^' 

as  follows :  the  time  rate  of  change  of  the  moment  of  moment^im 
of  a  particle  about  an  axis  is  equal  to  the  moment  about  the  axis 
of  the  forces  acting  upon  it. 

Compare  this  equation  and  its  interpretation  with  the  equation 
ini(d-Xi/dt^)=f^,  which  may  be  written  d{i)iiV-^)/dt=f^.  The  quantity 
iriiVi  is  called  the  momentum  of  the  particle  7h,-;  the  equation  states  that 
the  time  rate  of  change  of  the  momentum  in  any  direction  is  equal  to  the 
total  force  acting  in  that  direction. 

If  we  add  the  equations  like  the  above  for  all  masses,  the  inter- 
nal forces  disappear  as  above  explained,  and  writing  the  analogous 
equations  for  the  other  axes,  we  have 

--^^Y,,.^i_,,.^\  =  j^L  (3) 


-^dt     'y  dt  '  dt 

v-v  f ^       /    dv;         dx\ 

>  —  ml  X-  — ^  —  v-  — - 1  =  iv 

-^dt     \  '  dt      ^'  dt) 

where  L,  M,  and  N  are  the  sums  of  the  moments  about  the  axes 
of  all  the  external  forces  acting  upon  the  body.  These,  with  equa- 
tions (2),  form  the  required  six  differential  equations.  The  last 
three  (3)  have  the  disadvantage,  however,  that  they  do  not  contain 
a  limited  number  of  coordinates  which  simply  fix  the  position  of 
the  body  in  case  it  is  rigid.  This  cannot  be  satisfactorily  done 
without  introducing  moving  coordinate  systems,  which  wovdd  lead 
us  too  far  at  present,  although  it  would  furnish  a  basis  for  con- 
sidering most  interesting  motions  like  those  of  rotating  planets,  or 
of  g}'roscopes  and  "tops,"  the  latter  being  the  name  applied  in 
mathematical  physics  to  any  rigid  body  with  one  point  fixed.  It 
is  the  problem  which  naturally  follows  the  compound  pendulum, 
which  has  two  points,  or  an  axis,  fixed. 


ANALYSIS  107 

We  close  by  pointing  out  the  results  which  obtain  where  no 
external  forces  are  present.  Equations  (2)  and  (3)  admit  of  an 
integration  at  once,  the  first  set  giving  the  result  valid  for  a  rigid 
body  or  for  a  system  of  particles :  the  center  of  mass  moves  in  a 
straight  liiu  with  constant  velocity ;  and  the  second,  the  sum  of  the 
moments  of  momentum  about  any  fixed  axis  is  constant. 

These  results  find  interesting  application  in  our  solar  system. 
The  student  who  is  interested  in  a  further  study  of  the  motion  of 
rigid  bodies  is  referred  to  more  extended  works  on  dynamics.* 

ANALYSIS  OF  CHAPTER  V 

1.  Definitions  of  displacement,  translation,  and  rotation. 

2.  The  instantaneous  motion  of  a  rigid  body  consists  in  an  instantane- 
ous translation  and  an  instantaneous  rotation. 

3.  Assumptions  concerning  internal  forces. 

4.  The  work  and  kinetic  energy  for  a  motion  of  translation. 

5.  The  work  and  kinetic  energy  for  a  motion  of  rotation. 

6.  Definition  of  moments  and  products  of  inertia  and  the  radius  of 
gyration. 

7.  Two  theorems  enabling  us  to  get  the  moment  of  inertia  about  any 
axis  from  the  coeflBcients  of  inertia. 

8.  The  work  and  kinetic  energy  for  the  general  motion  of  a  rigid  body 
may  be  considered  the  sum  of  the  work  and  kinetic  energy,  respectively, 
due  to  a  translation  plus  the  work  and  kinetic  energy  due  to  a  rotation. 

9.  Compound  pendulum.  Experimental  determination  of  moments  of 
inertia.    Interchangeability  of  point  of  suspension  and  center  of  oscillation. 

10.  General  equations  of  motion  of  a  rigid  body.  The  student  should 
gain  an  idea  of  the  nature  of  the  problem  as  one  of  reducing  the  number 
of  variables  from  an  unlimited  number  to  six. 

*  For  instance,  Jeans,  Theoretical  Mechanics ;  AVebster,  Dynamics. 


SUGGESTIONS  AND  ANSWERS 

Note.  The  numerical  results  following  are,  for  the  most  part,  computed 
with  a  30-cm.  slide  rule.  The  student  should  therefore  not  expect  agreement 
beyond  |%. 

Page  4,  Ex.  1.    Use  Fig.  1. 

Page  5,  Ex.  3.  Draw  a  diagram,  indicating  the  sum  vector.  Xext 
show  that  the  order  of  any  consecutive  two  of  the  given  vectors 
may  be  changed  without  affecting  the  result.  The  theorem  will 
then  follow  if  it  can  be  shown  that  any  order  may  be  made  any 
other  order  by  such  interchanges  of  consecutive  vectors. 

Pages  9-11.  n.  Problems  on  Vectors.  3.  V2  —  V2/=  1.85/.  4.  v  = 
\/(734o  -  3408  V2)  =  50.3  ft.  per  sec;  47.5°  E.  of  X.  8  (a).  (16,  6, 
6);  (d)  (2,  9.83,  -  .869).  9  (a).  2.38,  (.728,  -  .485,  -.485);  (b)  0, 
(cosines  indeterminate).  10.  22.9  ft.  per  sec;  11.3°  with  track. 
11.  Against  current  at  angle  arccos(a/^)  with  bank.  If  a  =  h,  im- 
possible. Shortest  time,  90°  with  bank ;  path  then  makes  an  angle 
arctan(J/a)  with  bank.    12.  53°  from  the  vertical.    13.  16.7°.  v 

Page  13,  Ex.    May  be  reduced  in  part  to  Ex.  3,  p.  5. 

Pages  13-16.  III.  Problems  on  Equilibrium  of  Concurrent  Forces. 
2.  Consider  force  polygon.  4.  Use  conditions  (1)  and  Ex.  1,  p.  4. 
37.7,  (—  .647,  —  .727,  —  .262).  7.  Let  x,  y,  z  be  the  coordinates  of 
Oj  and  Xi,  y^,  Zi,  x^,  y^,  «2>  •  •  •,  ic„,  2/nj  ^n  ^^^  coordinates  of  the  other 
points.  Express  in  terms  of  these  the  projections  of  the  vectors  OA, 
OB,  •  •  •,  ON.  8.  The  components  directly  up  the  plane  of  the  forces 
must  be  in  equilibrium.  9.  As  a  check  note  that  the  sum  of  the 
angles  should  be  360°.  10.  52  lb;  11.  Lame's  theorem  maybe  used 
(problem  6);  84  and  112  lb.  12.  The  chains  are  to  be  considered  as  of 
equal  length.  70.7  lb.  13.  Solve  a  trigonometric  equation.  Get  all 
roots.  15.  .5  lb.  17.  The  graph  is  a  broken  straight  line.  18.  The 
graph  is  partly  curved  and  partly  straight.  19.  /n  =  .3.  The  reac- 
tion of  the  plane  is  equal  and  opposite  to  the  other  forces  acting 
upon  the  mass.  10.4  lb.  16.7°  with  vertical.  21.  t  =  arctan3/x. 
22.  Height  above  bottom  of  bowl :  r (1  —  cos  e)  =  ?•  (1  —  1  /  VT+jn^). 

109 


110  SUGGESTIONS  AND  ANSWERS 

Page  18,  Ex.  4.  Interpret  the  vanishing  of  the  product /„-^>  =  0, 
supposing  f4^  0. 

Page  19,  Ex.  1.  This  may  be  done  by  finding  the  moment  with 
respect  to  any  point  {x^  y)  and  showing  that  x  and  y  do  not  enter  the 
result.  Take  a  simple  position  of  the  axes  with  respect  to  the  forces. 
Ex.  2.  Apply  the  preceding  exercise  and  Ex.  2,  p.  18. 

Page  22,  Ex.  3.  First  show  that  if  a  set  of  forces  is  in  equilib- 
rium, so  also  are  their  projections  upon  any  plane ;  do  this  by  con- 
sidering the  resultant  and  also  the  moment  about  any  perpendicular 
to  the  plane.  Then  show  that  three  forces  in  a  plane  which  are  in 
equilibrium  must  be  concurrent  or  parallel ;  do  this  by  considering 
moments  about  the  intersection  of  any  two  of  the  forces  if  any  two 
intersect.  Considering  then  the  projections  of  the  three  given  forces 
upon  the  three  coordinate  planes,  show  that  the  forces  must  be  con- 
current or  parallel,  thus  proving  (b).  To  prove  (a)  apply  the  condi- 
tion for  equilibrium  of  concurrent  forces  (p.  13),  or,  if  they  are 
parallel,  consider  moments  about  a  line  meeting  and  perpendicular 
to  two  of  them.  Considering  then  the  resultant  of  parallel  forces 
through  the  origin,  (c)  may  be  proved. 

Pages  23-26.  IV.  Problems  on  Moments  and  the  Equilibrium  of 
Forces.  2.  The  resultant  of  the  parallel  forces  is  15  lb.  Consider- 
ing moments  about  the  lighter  end  of  the  bar,  the  resultant  is  found 
to  be  applied  at  a  distance  3|  ft.  from  this  end.  3.  1\\  ft.  from  the 
5-lb.  weight.  4.  Call  the  lengths  of  the  arms  d  and  cV  and  the  weight 
of  the  body  w.  Writing  the  two  equations  expressing  equality  of 
moments  and  solving  for  w,  one  finds  8.48  oz.  5  (a).  6  T.;  (b)  5|  and 
61  T.  6.  r  =  3  lb.,  R  =  2\  lb.,  *  =  0.  7.  45°.  8.  The  forces  acting 
upon  the  gate  are  its  weight  and  the  reactions  of  the  hinges,  which 
may  have  various  directions.  9. /=  w(^  —  ?-tan /)cot  i*/2  Z  ;  the 
center  is  directly  above  the  point  of  contact  with  the  floor.  10.  If 
the  triangle  of  the  rod  and  two  strings  be  thought  of  as  rigid,  there 
will  be  seen  to  be  two  forces  acting  upon  it,  the  upward  pull  of  the 
hook  and  the  downward  pull  of  gravity  upon  the  rod  considered  as 
acting  at  the  mid-point.  Hence  the  projections  of  a  and  i  on  a  hor- 
izontal line  may  be  shown  to  be  equal.  Then  project  upon  a  hori- 
zontal line  the  tensions  of  the  strings  and  the  weight  of  the  bar. 
11.   T==2cw/l;  P  =  w  Vl  -  2  c/^  -I-  (2  c/lf. 


SUGGESTIONS  AND  ANSWERS  HI 

Page  28.  V.  Problems  on  the  Center  of  Mass.  1.  10  in.  from  the 
heavy  end. 

Page  29,  Ex.  Consider  whether  the  formulas  of  p.  27  hold  for 
oblique  axes ;  then  consider  the  relation  of  the  product  Aa;A?/A«  to 
the  volume  of  the  parallelopiped  with  sides  parallel  with  the  axes 
and  of  lengths  A.x,  A?/,  and  As;. 

Pages  30,  36-40.  VI.  Problems  on  Centers  of  Mass  of  Continuous 
Bodies.  2.  On  the  line  joining  a  vertex  to  the  medial  point  of  the 
opposite  face  triangle  and  three  fourths  the  distance  from  the  vertex. 
3.  3  r/8  from  center,  4.  Check  by  noting  that  when  a  =  b  =  c  =  r 
the  ellipsoid  becomes  the  sphere  of  the  preceding  problem,  5.  2  h/3. 
7(a).  21/3;  (e)  check  by  considerations  of  symmetry;  (f).636^. 
S.x  =  .425  a,  y  =  A2o  b.  10  (a).  One  third  the  way  along  the  diagonal 
(the  plates  should,  of  course,  be  thought  of  as  concentrated  at  their 
centers  of  mass) ;  (b)  one  third  the  length  of  the  side  from  the  middle 
plate.  11..  Consider  either  as  a  conical  surface  or  as  a  solid  cone.  The 
result  may  be  obtained  from  the  answer  to  problem  7  (c).  13.  At  cen- 
ter of  axis.  14.  See  problem  7(d).  16.  y  =  .393.  19.  r/2  from  center, 
2Q.2{a^  +  ab  +  b'^)/3{a  +  b).  22.S  =  -4«/5.  2Z.x  =  {a^/mhym{h / 
am),  y  =  (a'^/vih)  [1  —  cos  (h/am)^,  z  =  h/2.  24.  x  =  C /M,  y  = 
S/M,  where  C  =2a((^''--l)/(l+4a^),  S  =  -(e*''' -1)/(1 +4.a''), 
M={f''^-V)la.  25.  3  A /8  from  one  end.  1^.  3iW ^bhi^ba'  +  a'') 
/^(Jy^  -\-ba  +  a"^).  28.  x  =  2r  sin%/3  {a  —  sin  a  cos  a).  Check  by 
noting  that  as  a  =  0,  a;  ^i^  r.  30.  x  =(3  a^  -\- IQ  ab  -{- 3  b^)/5  (a  +  b). 
31.  .652  a.  32.  X  =-  ar'/2(ab  -  r^).  34.  y  =  4a/3.  36.  x  = 
-5«/G.  37.  x  =  .555«.  38,  ^  =  . 388a.  ^Q.x  =  y  =  2a/5.  41.  r/7r 
from  axis,  h/4:  from  base.  42.  Fig.  14.  Distance  from  left-hand 
edge  =(2  a""  +  be  -  2  c^) /2(2  a  +  b  -  2 c);  Fig.  15.  Distance  from 
bottom  =  (ac'  -f  2  abc  +  b^d)/2  (ac  +  b(T) ;  Eig.  16.  y  =  5.41.  43.  x 
=  .538  a,  y  =  1.199 a.  44.S  =  8a./15,  y  =  152a/525,  «  =  11  a/120. 
45.  x  =  b^C/3M,  ?/  =  (5.\S/3il/ where  C=[e3«"'(sina-|-3mcosa)— 3m]/ 
(1  -f  9  m^),  S  =  [e3'""(3  m  sin  a  -  cos  a)  -f  1]/(1  -h  9  m"),  M  =  (^.y 
4  m)  (e^™"  —  !)•  46.  *418  a.  47.  Distance  from  edge  =  3  7rr  sin  a/ 
16  a.  Check  by  putting  a  =  7r/2,  and  also  a  =  7r;  cf,  problem  14, 
50.  A  =  4  ir^ab,  V=2  7rhi%.  51.  5  =  3  r(l  -|-  cos  a)/8.  52.  In  this 
problem  the  mass  in  question  is  understood  to  be  ojitside  the  con- 
ical surface  as  opposed  to  problem  51.     x  =  —  3r(l  —  cos  a)  / 8. 


112  SUGGESTIONS  AND  ANSWERS 


53.  X  =  3  r  sinVt/4(2  —  2  cos  a  —  siii^a  cos  a).    54.  x  =  3  Vr^  — a^/S. 

yxdx.  Apply  to  half  the  plate  to  the  left  of  a;  =  ira, 

which  half  will  have  the  same  y  as  the  whole  plate,  x  =  ira,  ]/  =  5  a/6. 
57.  Use  the  integration  tables.  S  =  a  [6  V2  -  log,(3  +  2  V2)]  / 
8  [  V2  +  log,(l  +  V2)]  =  .365  a.  58.  r>y  "  cone  "  is  meant  conical 
surface,  and  all  matter  is  removed  from  the  sphere  wliich  lies  within 
the  indefinitely  prolonged  conical  surface,    x  =  r  sin^a. 

Pages  46-48.  VII.  Problems  on  Rectilinear  Motion.  2(a).  t  =  —l, 
s  =  G;  (b)  points  to  left  of  *-  =  6 ;  (c)  forward  to  ^  =  G,  then  back ; 
(d)  part  of  line  to  left  of  a-  =  6,  twice ;  (e)  to  the  left ;  (f)  for  large 
negative  t,  the  point  is  far  to  the  left  with  large  positive  but  dimin- 
ishing velocity  ;  for  large  positive  t,  the  point  is  far  to  the  left  with 
large  negative  and  numerically  increasing  velocity.  20,  21.  The 
equations  give  more  than  one  value  of  s  for  a  given  value  of  t.  "What 
would  this  mean  ?    22.  s  =  (kt^  +  2)/2.    24.  s  =  2(e''  +  k-  2)/k. 

Pages  52-54.  VIII.  Problems  on  Bodies  moving  in  Straight  Lines 
under  the  Action  of  Given  Forces.  2.  403.6  ft.  5.  Determine  the  height 
h  of  the  body  at  its  highest  point  and  thus  answer  the  problem. 
6.  53  ft.  7.  The  weight  of  only  one  of  the  bodies  is  effective  in  pro- 
ducing motion,  while  the  inertia  of  both  bodies  resists  the  motion. 
They  are,  of  course,  supposed  to  start  from  rest,  (c)  v  =  242  cm.  per 
sec,  8.  AVrite  an  equation  for  each  body  in  which  its  acceleration  is 
equated  to  its  mass  times  the  resultant  of  the  forces  acting  upon  it, 
namely  gravity  and  the  tension  of  the  cord.  From  these  equations 
determine  the  unknown  acceleration.  10.  About  3  ft.  13.  The  path 
of  the  particle  must  bisect  the  angle  between  the  vertical  and  the 
perpendicular  to  the  line  on  to  which  it  falls.  14.  About  150  lb. 
15.  Six  times  as  high.  5.18  sec.  16.  About  10.3°.  17.  6012  ft. 
18.  410  ft.  19.  5.7  T.,  or  about  57  lb.  per  T.  20.  About  1268  sec, 
or  21.1  min.  25,950  ft.  per  sec.  22.  1  ft.  23.  About  7.8  kgm. 
24.  25,950  ft.  per  sec.  28.1  min.  26.  k  =  17.6 ;  10.15  ft.  per  sec. 
27.  Velocity  of  bullet  is  about  3.4  times  that  of  raindrop.  30.  Veloc- 
ity upon  leaving  table  is  V5/y  log^(5  +  V24)  =  .904  ft.  per  sec. 

Pages  73-76.  X.  Problems  on  Bodies  moving  in  a  Plane  under  the 
Action  of  Forces.  2.  106  ft.  per  sec.  4.  6.2°.  6.  2.63  ft.  10.3°  from 
vertical.  7.  Solve  the  problem  from  the  standpoint  that  the  stream 
of  water  is  sent  from  a  fixed  point,  and  consider  the  area  above  the 


SUGGESTIONS  AND  ANSWERS  113 

level  of  that  point  only.  1100  sq.  m.  9.  Use  equation  of  envelope. 
16.  By  ^Ijj  of  its  length.  19.  The  centrifugal  force  is  viv'^/p 
poundals,  if  m  is  measured  in  pounds.  1.34  lb.  20.  433  T.  22. 14.9°. 
23.  About  1  /291  and  17.  24.  The  data  lead  to  a  result  45  ft.  As  this 
is  a  very  short  throw,  how  should  the  data  be  changed  ?  25.  4  a^mv^ J 
{if  -f  4  «"-^)i  poundals.     28.  y  =  ?/o  +  (kh  —  g)/k{l  —  cos  VJct). 

Pages  78,  79.  XI.  Problems  on  Work  and  Power.  3.  37.9  h.  p. 
5.  .14  ft.  lb.  6.  5ih.  p.  9.  800  h.  p.  10.  {b  +  a-2l){b  -  a)X/2l 
ft.  lb.     11.  1.21  h.  p. 

Page  84.  XIII.  Problems  on  Energy.  (Notk.  In  these  problems 
the  student  must  be  careful  about  units.  In  the  derivation  of  the 
formula  for  k  =  ^viv"^,  p.  83,  the  equations  f^.  =  vi{dv^/dt),  f^  = 
vi(dVy/dt)  hold  if  m  is  measured  in  pounds  and/^  3,nd/j,  inj^oundals; 
and  also  if  m  is  measured  in  engineering  units  (weight  -f-  (/)  and  f^ 
and/y  in  pounds.  Thus  if  7n  is  measured  in  pounds,  k  =  ^mv^foot 
poundals ;  if  m  is  measured  in  engineering  units,  /c  =  J  vfiv^  foot 
pounds.)  2.  19,100  ft.  lb.  3.  368,000  ft.  lb.  4.  To  a  point  24  ft. 
higher  than  the  foot  of  the  hill.  6.  4'^.  7.  v^/2g  ft.  above  the 
lowest  point  if  the  rod  is  long  enough.  10.  5850  h.  p.  11.  At  a 
height  2  a/3  above  the  level  of  the  center. 

Pages  95-98.  XIV.  Problems  on  Moments  and  Products  of  Inertia 
and  Radii  of  Gyration.  5.  /712  ;  /V24;  l^/Z.  6.  A  =B=  iyi2,  C= 
l*/Q,  D  =  E  =  F  =  0;  I,  =  l*/12;  I,  =  2iy3;  I,„  =  5l'/12.  S,A  =  B 
=  7rr*/4:,  C  =  7rr*/2;  r  =  l/</S7r  =  .57l.  9.r  =  .63L  n.C=2ar^. 
12.  A  =  TrahyA,  C  =  irab  (a^  +  b"")  /4  ;  /^-,„  =  irab  (4  a""  -  3  b^)  /  A  • 
Ij.^  =  irab  (5  a''  -  3  b'')/A.  13.  If  A  =f(a,  b,  c),  then  B  =f(b,  c,  a), 
C=f(c,  a,  b).  A=A  -rrabc  {V"  +  c^)  /15.  15.  C  =  35  7raV16,  .4  = 
21  7raV32.  16.  A  =  abc  (b^  +  ^•')  /12  ;  4  =  abc  (b^  +  c^)/3;  /,/  = 
abc  {a%^  +  ^'^2  +  c^a^)  / 6  (a^  +  ^2  +  c^).  1 7.  abc  [2  «^ {b^  +  f;^)  +  ^,2^2]  / 
6(J2  -I-  c2).  19.  C  =  7ra?-y  10.  The  result  of  problem  8  may  be  used 
to  advantage  here.  20.  A^  B  =  'Trar^(a^  +  3  r^)/12,  C  =  7rr^a/2. 
21.  A  =  43^,  B  =  79r;,   C  =  A  +B  =  123,  F  =  -  15.     25.  r'  =  .235  r. 

Pages  101-102.  XV.  Problems  on  Work  and  Energy  in  the  Case  of 
Rigid  Bodies.  6.  A'^ :  A'<  =  2  :  1  for  cylinder  and  5  :  2  for  sphere. 
7.  Difference  in  heights  of  centers  of  spheres  is  10(a  +  r)/17  at 
time  of  leaving. 


INDEX 


[The  numbers  used  below  refer  to  pages.] 


Acceleration,  2,  8,  42,  55 
Angle  of  friction,  16 
Angular  acceleration,  6-4 
Angular  velocity,  64 
Areas,  integral  oi,  70 
Arm  of  a  force,  17 

Bodies  of  revolution,  center  of  mass 
of,  31 

Center  of  mass,  26,  105,  107 
Center  of  oscillation,  103 
Central  forces,  68 
Central  motion,  68,  69 
Circular  motion,  63 
Coefficient  of  friction,  16 
Coefficients  of  inertia,  93 
"Complete  description"  of  a  motion, 

42,  55 
Composition  of  forces,  5,  22 
Concurrent  forces,  concurrent  vectors, 

6,  12 
Conditions  for  equilibrium,  13 
Conservation  of  energy,  71,  83 
Conservative  fields,  80,  100 
Coordinates  of  center  of  mass,  27,  29 
Couples,  18,  19 
Cylinders,  center  of  mass  of,  30 

Density,  28 

Differential  equations,  of  rectilinear 
motion,  43,  44,  45 ;  of  plane  motion, 
63 ;  of  motion  of  a  rigid  body,  104 

Displacement,  86 

Dynamical  equivalence,  95 

Dynamics,  40 

Energy.    See  Potential  energy,  etc. 
Energy  integral,  71 
Energy  of  a  rigid  body,  100 
Equilibrium,    12,   82  ;    conditions  for 

(rigid  body),  20,  22  ;  of  concurrent 

forces,  12 
Equipotential  surfaces,  81 


Fall  from  a  great  height,  49 

Force.  See  Units ;  also  Central  forces, 

etc. 
Force  fields,  79 
Force  polygon,  13 
Friction,  15 

Graphic  representations  of  a  motion, 

45 
Gravitation,  49 
Gravitational  units.   See  Units 

Harmonic  motion,  simple,  48,  69 ; 
elliptic,  69 

Inclined  plane,  14,  1.5,  16,  48 
Inertia.   See  Moment  of  inertia,  etc. 
Listantaneous  motion,  86 
Integral.   See  Areas,  integral  of,  etc. 
Internal  forces,  89 

Kepler's  law,  73 

Kinetic  energy,  71,  82,  89,  100 

Lamp's  theorem,  14 
Lissajou's  curves,  76 

Mass.  See  Units ;  also  Center  of  mass, 
etc. 

Mechanics,  1 

Moment  of  a  couple,  19 

Moment  of  a  force,  16,  17,  18 

Moment  of  a  velocity,  105 

Moment  of  inertia,  91,  92;  experi- 
mental determination  of,  103 

Moment  of  momentum,  106,  107 

Momentum,  106,  107 

Motion.  See  Rectilinear  motion.  Plane 
motion,  etc. 

Newton's  law  of  gravitation,  49 
Newton's  third  law,  89 
Nonconcurrent  forces,  16 


115 


116 


ESTDEX 


Normal  acceleration,  57,  59,  64 
Normal  velocity,  57 

Oscillation,  center  of,  103 

Pappus,  theorems  of,  40 

Particle,  26 

Pendulum,   66,  67,  68,  75,  76; 

pound, 102 
Perpetual  motion,  83 
Plane  motion,  55,  63 
Planetary  motion,  71 
Plates,  31 

Potential  energy,  71,  80,  89 
Power,  78 

Products  of  inertia,  93 
Projectile  motion,  65,  66,  74,  75 

Radius  of  curvature,  59 
Radius  of  gyration,  95,  103 
Rectilinear  motion,  42 
Resisting  medium,  50 
Resultant  of  forces,  5,  22 
Rigid  bodies,  12,  86,  98 


Rotation,  86,  90 

Simple  harmonic  motion,  48,  65,  69 

Speed,  56 

Statics,  12 

Surfaces  of  revolution,  center  of  mass 

of,  33 
Surfaces,  equipotential,  81 

Tangential  acceleration,  57,  58,  59,  64 
Tangential  velocity,  57,  58,  59,  64 
Translation,  86,  89 

Uniform  motion,  42,  48 
Units,  1,  78 

Vectors,  3 ;  composition  of,  5,  22 
concurrent,  6,  12  ;  derivatives  of,  7 
fixing  of,  3 ;  operations  upon,  4 
polygon  of,  13  ;  resolution  of,  5 

Velocity,  42,  64 

Wires,  center  of  mass  of,  31,  33 
Work,  78  ;  rigid  body,  89,  91,  100 


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By  S.  E.  Slocum,  Assistant  Professor  of  Mathematics  in  the  University  of  Illinois,  and 
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\     COLLECTION  of  over  five  hundred  practical  problems  is  here 
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William  F.  Allison,  Professor  of  Civil  Engineering,  Colorado 
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